The speed by which the spaceprobe travels when all the rocket fuel is used up will be 29.262 m/sec.
What is the law of conservation of linear momentum?
According to the law of conservation of linear momentum before the collision is equal to the momentum after the collision. These laws state how momentum gets conserved.
Unit conversion;
1 km/sec = 1000 m/sec
Given data;
Spaceprobe speed = 1.795 km/s = 1795 m /sec
Probe mass = 635.0 kg
Fuel mass = 4092.0 kg
Expelled propellent velocity = 4.161 km/s = 41461 m/sec
From the momentum conservation principle;
[tex]\rm P_i = P_f \\\\ (m_p+m_f)v_i = m_pV - m_fv_p \\\\ V = \frac{(635+4092)1795+4092 \times 41461}{635} \\\\ V = 280540.7 \ m/sec \\\\ V = 28.05 m/sec[/tex]
Hence, the speed by which the spaceprobe travels when all the rocket fuel is used up will be 29.262 m/sec.
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2. Two small (green) objects, each of mass m, are separated by a solid, massless rod of length L. They are located so that one of the objects is located at a distance of r away from the center of a uniform spherical planet with mass M (see figure). M r m L m Assume that m is very small so that you can ignore the gravitational force between the two small (green) objects. Is the rod being compressed, stretched, or neither? If compressed/stretched, calculate the magnitude of this deformation force on the rod. If neither, explain clearly why.
The rod is stretched due to the force of gravitational attraction.
The magnitude of the deformation force on the rod is -GMm×((1/r² + 1/(L+r)²)
What is gravity?The force of attraction felt by a person which is directed at the center of a planet or Earth is called as the gravity.
The force of attraction is directly proportional to the product of masses of the object and inversely proportional to the square of distance between them.
F = GMm/R²
Given are two small (green) objects, each of mass m, are separated by a solid, massless rod of length L. They are located so that one of the objects is located at a distance of r away from the center of a uniform spherical planet with mass M. Assume that m is very small so that you can ignore the gravitational force between the two small (green) objects.
Let the mass closer to the planet is A and the other is B
FA = Force on planet A = -GMm/r²
FB = Force on planet B = -GMm/(L+r)²
Net Force on the rod is given by the addition of the individual forces.
Fnet= FA + FB = -GMm/r² -GMm/(L+r)²
Fnet = -GMm×((1/r² + 1/(L+r)²)
Thus, the magnitude of the deformation force on the rod is derived above.
A experience more force than B, so A will stretch out from B. Hence the force is stretching the rod.
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143°C = _____
416 K
-130 K
0 K
143 K
28. An electron with a speed of 4.0 x 10° m/s enters a uniform magnetic field of magnitude 0.040 T at an angle of 35 degrees to the magnetic field lines. The electron will follow a helical path. a) Determine the radius of the helical path. b) How far forward will the electron have moved after completing one circle?
Answer:
r= 1.09×10^-4
Explanation:
Given
V=speed=4.0×10^5 m/s
B= magnetic field= 0.040 T
©=angle= 35°
m= mass of electron= 9.11×10^-31
q= charge of electron= 1.60×10^-19
solution
qv×B= mv²/r
qvBsin©=mv²/r
qBsin©=mv/r
r=mv/qBsin©
r=9.11×10^-31× 4.0×10^5/1.064×10^-19×0.04T(sin35°)
r= 1.09×10^-4 m
a) r = 1.09 * [tex]10^{-4}[/tex] m
b) Distance travelled : 6.845 * [tex]10^{-4}[/tex] m
What is an electron ?
An electron is a stable subatomic particle with a charge of negative electricity, found in all atoms and acting as the primary carrier of electricity in solids.
given
charge of electron : 1.6 * [tex]10^{-19}[/tex] C
mass of electron = 9.11 * [tex]10^{-31}[/tex] kg
v = 4.0 x 10 m/s
B = 0.040 T
theta = 35 degrees
since ,
force in magnetic field on electron = centripetal force
a) q(v*B) = m [tex]v^{2}[/tex] / r
q v B sin(theta) = m [tex]v^{2}[/tex] / r
r =m v /q B sin(theta)
r = 9.11 * [tex]10^{-31}[/tex] * 4.0 x [tex]10^{5}[/tex]/ 1.6 * [tex]10^{-19}[/tex] sin (35)
r = 1.09 * [tex]10^{-4}[/tex] m
b) far forward will the electron have moved after completing one circle will be equal to circumference of the circle = 2πr
= 2 * 3.14 * 1.09 * [tex]10^{-4}[/tex] m = 6.845 * [tex]10^{-4}[/tex] m
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please help me in this questions
Answer:
Rainy daywindy daysunny daycloudy daysorry I don't know the answer of question 8.sunglassesumbrella:-(:-):-):-(day ☀️night day ☀️night Day ☀️Day ☀️night nightExplanation:
Hope I give all correct answer please mark as brainlest answer
An ultraviolet wave traveling through a vacuum has wavelength of 4.0 x 10^-7 m. The waves frequency, written in scientific notation to two significant figures, is ? X10^14Hz.
Answer:
λ = c / f or f = c / λ
f = 3.0E8 / 4.0E-7 = .75E15 / sec = 7.5E14 / sec = 7.5 X 10^14 /sec
an object is 27.0 cm from a concave mirror of focal length 15.0 cm. find the image distance.
The distance of the Image will be -33.75 cm
A concave mirror has an inward-curving reflecting surface that faces away from the light source. Unlike convex mirrors, a concave mirror's image forms a variety of images based on the object's proximity to the mirror.
Given that, an object placed 27 cm from a concave mirror having the focal length of 15 cm
We have to find distance of the Image
Using Mirror Formula:
1/f = 1/v + 1/u
Where,
f = focal length
v = Image distance from the mirror
u = object distance from the mirror (concave)
Substitute the known values in the above formula to find the value of 'v' i.e. from the mirror.
1/(-15) = 1/v + 1/(-27)
1/(-15) = 1/v - (1/27)
1/v = -0.029
v = -33.75 cm
Therefore the distance of the Image will be -33.75 cm
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water
0.6m
water wave
tank
How long does it take for the wave to return to the
position XY, but moving to the right?
[3]
b A man is cutting down a tree with an axe. He
hears the echo of the impact of the axe hitting
the tree after 1.6 s.
i What sort of obstacle could have caused the
echo?
ii The speed of sound is 330 m/s. How far is
the tree from the obstacle?
c Distinguish between the nature of the sound
wave in b and the water wave in a.
[2]
ii the amplitude of
b The cone of a louds
diagram shows how
out in front of the c
loudspeaker
P is a compression,
i
Describe how
changes from
ii Describe the
the sound w
iii Copy the di
and mark a
wavelength
5 a The first diagrams
Answer:
Discrimination is the worst thing in the world you can't even do a thing so you will do such a physical thing or do a mathematics problem ok done it's ok
The road from city A to city B is described by a car with Vm 40 km / h. When the car turns (from B to A) the average speed is 60 km / h. Find the average round trip speed.
Answer:
i think the answer is 20.......
'The wave' is a particular type of pulse that can propagate through a large crowd gathered at a sports arena to watch a soccer, hockey, or CFL game. The elemets of the medium are the spectators, with zero position corresponding to their being seated and maximum position corresponding to their standing and raising their arms. When a large fraction of the spectators participate in the wave motion, a somewhat stable pulse shape can develop. The wave speed depends on people's reaction time, which is typically on the order of 0,1s. Estimate the order of magnitude, in minutes, of the time required for such a pulse to make one circuit around BC Place Stadium in Vancouver.
State all the assumptions that you've made.
Information about BC Place: dimensions are approximately 100 m X 85 m.
The total time required is 1 minute and 0.3 seconds.
Assumption: The distance between the people is 1 m and the stadium is a circle with a radius of 100m.
Here, the time taken by the person is 0.1 seconds.
Total distance covered by the wave = Circumference of the circle
So, total distance = 2 πr = 2 × 3.14 × 100 = 618 m
As the distance between each of the people is 1 m.
So, the number of personal interactions is 618.
Time taken by each person is 0.1 seconds.
So, total time, t = 0.1 × 618 = 61.8 seconds.
So, the order of the magnitude of the time required is 1 minute and 0.3 seconds for such a pulse to make one circuit around BC Place Stadium in Vancouver.
Hence, the total time required is 1 minute and 0.3 seconds.
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After watching the video below and based on your personal experiences, is there a difference
between areas in precision of control? Could there be differences between left and right cortex
based on experience such as handedness or specific skills such as playing a guitar?
Based on my personal experiences, I believe that there is a difference between areas in precision of control because there are different part of the brain that controls the functions of the body.
What is the difference between left-handed and right-handed people?From the standpoint of brain lateralization, differences do exist such as based on experience such as handedness or specific skills such as playing a guitar.
Note that Left-handers are said to have reduced or little lateralized brains, which tells us that the two halves of the brain are little different than as seen in the right-handers.
Therefore, I can say that based on my personal experiences, I believe that there is a difference between areas in precision of control because there are different part of the brain that controls the functions of the body.
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What method could I use to test this hypothesis? If the mass and the volume of and object are known, then its density can be calculated dividing the object's mass by its volume.
Answer:
The scientific method
Explanation:
When passing in a medium for a distance of 1.5 cm intensity
of the light decreased by 3 times. What will the distance x equal to when the intensity of the light decreases by 9 times?
The distance x equal to 0.87 cm when the intensity of the light decreases by 9 times.
Intensity of light
The intensity of light is given as power emitted by the light by unit area.
I = P/A
I = P/L²
I₁L₁² = I₂L₂²
L₂² = I₁L₁²/I₂
L₂² = (3 x 1.5²)/(9)
L₂² = 0.75
L₂ = 0.87 cm
Thus, the distance x equal to 0.87 cm when the intensity of the light decreases by 9 times.
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If a 100 N block is resting on a steel table with a coefficient of
static friction μs = 0.68, then what minimum force is required to
move the block?
Answer:
6800
Explanation:
100 x 0.68=6800
Answer: 68
Explanation:
There are several ways to model a compound one type of model is shown ?what is the chemical formula for the molecule modeled?
The is organic compound with the correct chemical formula C4H9O2.
What is a model?A model is a representation of reality. A model serves the purpose of prediction as well as explanation.
Looking at the model of the molecule we can see that it is the organic compound with the correct chemical formula C4H9O2. The molecule is shown in the image attached to this answer.
Missing parts:
There are several ways to model a compound. One type of model is shown.
What is the chemical formula for the molecule represented by the model?
CHO
C4H9O2
C4H8O
C3H8O2
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Calculate the approximate number of atoms in a bacterium, assuming the average mass of an atom is ten times the mass of a hydrogen atom. The mass of a bacterium is 10−15 kg and the mass of a hydrogen atom is of the order of 10−27 kg.
atoms
10¹¹ is the approximate number of atoms in a bacterium.
What do you understand by mass of element?
The atomic mass of an element is the average mass of the atoms of an element measured in atomic mass unit (amu).
Given,
Mass of a bacterium atom = 10⁻¹⁵ kg.
Mass of a hydrogen atom = 10⁻²⁷ kg.
From the above observation ,
The average mass of an atom of the bacterium is ten times the mass of a hydrogen atom.
Atomic mass 1 bacterium atom = 10 x mass of hydrogen atom
= 10 x 10⁻²⁷ kg.
= 10⁻²⁶ kg.
Thus,
The number of atoms in a bacterium
= [tex]\frac{Total mass}{Atomicmass of 1 bacterium}[/tex]
= [tex]\frac{10^{-15} }{10^{-26} }[/tex]
= [tex]10^{11}[/tex]
The approximate number of atoms in a bacterium is [tex]10^{11}[/tex].
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Jose was out drinking with his friends for nearly the whole night. The next morning he was confused and vomiting, and had a low body temperature.
Answer:
He has a hangover.
Explanation:
Just something I know.
Taking the density of air to be 1.29 kg/m3, what is the magnitude of the angular momentum (in kg · m2/s) of a cubic meter of air moving with a wind speed of 73.0 mi/h in a hurricane? Assume the air is 51.2 km from the center of the hurricane "eye."
The magnitude of the angular momentum of air will be 4.128 x 10^(-3) kg·m^2/s
The rotating equivalent of linear momentum in physics is called angular momentum. Because it is a conserved quantity—the total angular momentum of a closed system stays constant—it is significant in physics. Both the direction and the amplitude of angular momentum are preserved.
Given the density of air to be 1.29 kg/m3 and a wind speed of 73.0 mi/h
We have to find the magnitude of the angular momentum
Let,
ρ = Density of air = 1.29 kg/m^3
v = Speed of wind = 73.0 mi/h = 0.032 km/s
M = angular momentum of air
Let the volume of air be 1 m^3
Mass = Volume x ρ = 1 x 1.29 = 1.29 kg
Momentum = M = mass x velocity
Momentum = 1.29 x 0.0032
Momentum = 4.128 x 10^(-3) kg·m^2/s
Hence the magnitude of the angular momentum of air will be 4.128 x 10^(-3) kg·m^2/s
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Calculate the estimated density of each ball. Use the formula
D = m/V
where D is the density, m is the mass, and V is the volume. Record your calculations in Table A of your Student Guide.
Given that the density of water is 1.0 g/cm3, make a prediction about whether each ball will float in water. Record your prediction in Table A.
What is the estimated density of the table tennis ball? Record your answer to the nearest hundredth.
0.07
g/cm3
What is the estimated density of the golf ball? Record your answer to the nearest hundreth.
The estimated density of the golf ball is 700 kg/m³
What is density?Density is defined as Mass per unit Volume.
In displacement method,
First , we measuring the volume of water displaced by an object which tell us the volume of the object then we will use the physical balance to determine its mass.
Then calculate the density by dividing the mass by the volume.
i.e. D = m/V
Given, Density of water is 1.0 g/cm³
Using displacement method , The estimated volume of golf ball is 100 cm³ and estimated mass is 7g
Then ,
Density = 10 cm³ / 7 g= 0.07 g/ cm³ = 700 kg/m³
So the estimated density of golf ball is 700 kg/m³
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A plane leaves with an acceleration of 6.34 m/s squared and takes 1.5 hours to stop. What is the speed of the plane? What was the distance it traveled?
The answer to this question is Initial velocity of plane will be 34236 m/s and 92437.2 Km is the distance travelled by it.
Three equation of motion are:-
v = u + ats = ut + (1/2)at²v² - u² = 2asWhere v is final velocity, u in initial velocity, s is the displacement by the object, a is the acceleration and t denotes the time.
In question we have given deceleration as 6.34 m/s² and time as 1.5 hour which is equal to 5400 seconds.
Applying equation 1 to find the initial speed of plane
v = u + at
0 = u + (-6.34 × 5400) {v=0 as plane will stop after 5400 sec}
u = 6.34 × 5400
u = 34236 m/s
Initial velocity of plane is 34236 m/s
Applying equation 2 to find the displacement of plane in that time period
s = ut + (1/2)at²
s = ( 34236 × 5400 ) - ( (1/2) × 6.34 × 5400² )
s = 5400 × ( 34236 - ((1/2) × 6.34 × 5400) )
s = 5400 × ( 34236 - 17118 )
s = 5400 × 17118 metres
s = 5.4 × 17118 Km
s = 92437.2 Km
Distance travelled by plane is 92437.2 Km
So, the initial velocity of plane will be 34236 m/s and the displacement of plane in that time period will be 92437.2 Km.
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Suppose the ski patrol lowers a rescue sled carrying an injured skier, with a combined mass of 97.5 kg, down a 60.0-degree slope at a constant speed, as shown in the Figure. The coefficient of kinetic friction between the sled and snow is 0.100.
a) How much work, in joules, is done by friction as the sled moved 28 m along the hill?
b) How much work, in joules, is done by the rope on the sled this distance?
c) What is the work, in joules done by the gravitational force on the sled?
d) What is the net work done on the sled, in joules?
a. 1337.3 J work, in joules, is done by friction as the sled moved 28 m along the hill.
b.21,835 J work, in joules, is done by the rope on the sled this distance.
c. 23,170 J the work, in joules done by the gravitational force on the sled d. The net work done on the sled, in joules is 43,670 J.
What is friction work?
The work done by friction is the force of friction times the distance traveled times the cosine of the angle between the friction force and displacement
a. How much work is done by friction as the sled moves 28m along the hill?
ans. We use the formula:
friction work = -µ.mg.dcosθ
= -0.100 * 97.5 kg * 9.8 m/s² * 28 m * cos 60
= -1337.3 J
-1337.3 J work, in joules, is done by friction as the sled moved 28 m along the hill.
b. How much work is done by the rope on the sled in this distance?
We use the formula:
Rope work = -m.g.d(sinθ - µcosθ)
rope work = - 97.5 kg * 9.8 m/s² * 28 m (sin 60 – 0.100 * cos 60)
= 26,754 (0.816)
= 21,835 J
21,835 J work, in joules, is done by the rope on the sled this distance.
c. What is the work done by the gravitational force on the sled?
By using the formula:
Gravity work = mgdsinθ
= 97.5 kg * 9.8 m/s² * 28 m * sin 60
= 23,170 J
23,170 J the work, in joules done by the gravitational force on the sled .
D. What is the total work done?
By adding all the values
work done = -1337.3 + 21,835 + 23,170
= 43,670 J
The net work done on the sled, in joules is 43,670 J.
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Consider the baby being weighed in Figure 4.25.
Figure 4.25
(a) What is the mass of the child and basket if a scale reading of 104 N is observed?
kg
(b) What is the tension T in the cord attaching the child to the scale?
N
(c) What is the tension T' in the cord attaching the scale to the ceiling, if the scale has a mass of 0.500 kg?
N
(d) Draw a sketch of the situation indicating the system of interest used to solve each part. The masses of the cords are negligible. (Do this on paper. Your instructor may ask you to turn in this work.)
The mass and tension due to the system are as follows:
The mass of the child and scale = 10.6 kgThe tension T, in the cord attaching the child to the scale = 104N The tension T', in the cord attaching the scale to the ceiling T' = 108.9 NWhat is tension?Tension is a type of pulling force due transmitted by means of a string or cable.
Force = mass * acceleration due to gravitya) The mass of the child and scale = 104/9.81 = 10.6 kg
b) The tension T, in the cord attaching the child to the scale = scale reading = 104N
c) The tension T', in the cord attaching the scale to the ceiling = scale reading + weight of scale
T' = 104 + (0.5 * 9.81)
T' = 108.9 N
d) The sketch is attached in the picture
In conclusion, the tension is force exerted on the cord due to the weight of the scale and the baby.
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Which of the following measurement is most significant?
A. 66.000cm
B. 0.00066cm
C. 6.600cm
D. 6.6cm
Option C. The measurement with the most significant number is 6.600 cm.
What is significant number?
Significant numbers are numbers that have significance or meaning and give more precise details about the value of the entire numbers.
66.000 cm ------> 2 significant numbers0.00066 cm -------> 2 significant numbers6.600 cm ----------> 4 significant numbers6.6 cm ---------------> 2 significant numbersThus, the measurement with the most significant number is 6.600 cm.
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Q3, A ball of mass 5.0 kg moving with a Velocity of 10.0 ms collides
with a 15.0 kg ball moves with a Velocity of 4 ms! If both balls
Stick together after Collision, Calculate their Common Velocity after Impact if they initially moves in The Same direction, and Opposite direction.
Answer:
Their common velocity after the collision will be 5.5m/s
Explanation:
look at the attachment above ☝️
Which picture correctly shows the path of refracted light rays given an object outside the focal point? Select one: a. A b. B c. C d. D
Answer:
Answer is C because light travels in a sight line but when light pass through a refractor the light from the source changes direction when passes through a refractor
Please help me!
If ball C is 3 times the volume of ball D and ball D has 1/3 the mass of ball C, which has the greater density?
A. Ball C
B. Ball D
C. The Densities are equal
C. The Densities are equal.
What is density?
Density is mass per unit volume or mass of a unit volume of a material substance.
If m1, V1 and D1 = mass, volume and density respectively of ball C
m2, V2 and D2 = mass, volume and density respectively of ball D
According to the Question ,
[tex]V_{1} = 3V_{2} , m_{2} = \frac{1}{3} (m_{1} ) \\ \\= m_{1} = 3m_{2}[/tex]
Therefore,
[tex]\frac{D_{1} }{D_{2} } = (\frac{m_{1} }{V_{1} } )* (\frac{m_{2} }{V_{2} } )\\ \\= (\frac{3m_{2} }{3V_{2} })*(\frac{V_{2} }{m_{2} }) \\\\= 1[/tex]
Hence, D1 = D2
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When compared to wave II, wave I represents a wave with...
Select one:
a. a higher frequency.
b. a lower frequency.
c. an equal frequency.
d. a greater amplitude.
Wave I stands for a wave with an equal frequency as wave II. Option c is correct.
What is the frequency?Frequency is defined as the number of repititions of a wave occurring waves in 1 second. Its unit is Hz.
Frequency is given by the formula as,
[tex]\rm f = \frac{1}{t}[/tex]
Where,
f is the frequency
t is the period of the wave
From the digrame it is observed that both the wave has the same period. So that they will have the same frequency.
Hence option c is correct.
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During a diversity management session, a manager suggests that stereotypes are a necessary part of working with others. “I have to make assumptions about what’s in the other person’s head, and stereotypes help me do that,” she explains. “It’s better to rely on stereotypes than to enter a working relationship with someone from another culture without any idea of what they believe in!” Discuss the merits of and problems with the manager’s statement.
The merits of and problems with the manager’s statement are explained below.
What is stereo typing in organizations?Extension of the social identity of the individual in groups such as working in company or studying in college is called as the stereotyping in organizations.
Merits of Stereotypes in organizations:
It relies on categorical thinkingHelps understand outside world easily'unique characteristics are difficult to recall every time.groups are identified easilyhelps filling the gaps while talking of some new culture.helps enhancing self perception and social identity.Demerits of Stereotypes:
difficulty in understanding behavior of other individual in organizations.an individual may be sometimes under or over estimated.discourages social group while entering a profession.discriminatory behavior.not helps in describing with their talent.Thus, the merits and demerits of stereotypes in organizations are explained above.
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A ball of mass 500g is dropped from a height 1.5m . It rebounds the floor and reached the height 1.2m. Calculate the impulse given to the bàll by the floor
The impulse given to the ball by the floor is 0.2865 kg.m/s.
What is impulse?The change in momentum is equal to the product of impact force applied while colliding and time for that impact.
Impulse F. t = m (Vf -Vi)
where, Vf is the final velocity and Vi is the initial velocity.
A ball of mass 500g is dropped from a height 1.5m . It rebounds the floor and reached the height 1.2m.
The initial velocity u = √2x 9.81 x 1.5 = 5.425 m/s
The final velocity v = √2x 9.81 x 1.2 = 4.852 m/s
Substitute the values into the expression, we get
Impulse = m(v- u)
Impulse=0.5 x (4.852- 5.425 )
Impulse = - 0.2865 kg.m/s
Thus, the magnitude of impulse given to the ball by the floor is 0.2865 kg.m/s.
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which has more KE, a 2 g bee flying at 1 m/s, or a 1 g wasp flying at 2 m/s
Answer:
the 1 gram wasp
Explanation:
To start off with this problem, write down every piece of information and do neccessary conversions.
mass of bee = 2 grams = 0.002 kg
speed of bee = 1 m/s
mass of wasp = 1 gram = 0.001 kg
speed of wasp = 2 m/s
now, we will use the kinetic energy formula and compare the answers
KE BEE = 0.5 (0.002 kg)(1 m/s)^2 = 0.001 Joules
KE WASP = 0.5(0.001 kg)(2 m/s)^2 = 0.002 Joules
0.002 J > 0.001 J
4. A ball is thrown with an initial speed vi at an angle θi with the horizontal. The
horizontal range of the ball is R, and the ball reaches a maximum height R
6
. In
terms of R and g, find (a) the time interval during which the ball is in motion,
(b) the ball’s speed at the peak of its path, (c) the initial vertical component of
its velocity, (d) its initial speed, and (e) the angle θi
Disclaimer: I ended up finding what's asked for in the reverse order (e)-(a).
At time [tex]t[/tex], the horizontal position [tex]x[/tex] and vertical position [tex]y[/tex] of the ball are given respectively by
[tex]x = v_i \cos(\theta_i) t[/tex]
[tex]y = v_i \sin(\theta_i) t - \dfrac g2 t^2[/tex]
and the horizontal velocity [tex]v_x[/tex] and vertical velocity [tex]v_y[/tex] are
[tex]v_x = v_i \cos(\theta_i)[/tex]
[tex]v_y = v_i \sin(\theta_i) - gt[/tex]
The ball reaches its maximum height with [tex]v_y=0[/tex]. At this point, the ball has zero vertical velocity. This happens when
[tex]v_i \sin(\theta_i) - gt = 0 \implies t = \dfrac{v_i \sin(\theta_i)}g[/tex]
which means
[tex]y = \dfrac R6 = v_i \sin(\theta_i) \times \dfrac{v_i \sin(\theta_i)}g - \dfrac g2 \left(\dfrac{v_i \sin(\theta_i)}g\right)^2 \\\\ \implies R = \dfrac{6{v_i}^2 \sin^2(\theta_i)}g - \dfrac{3{v_i}^2 \sin^2(\theta_i)}g \\\\ \implies R = \dfrac{3{v_i}^2 \sin^2(\theta_i)}g[/tex]
At the same time, the ball will have traveled half its horizontal range, so
[tex]x = \dfrac R2 = v_i \cos(\theta_i) \times \dfrac{v_i \sin(\theta_i)}g \\\\ \implies R = \dfrac{2{v_i}^2 \cos(\theta_i) \sin(\theta_i)}g[/tex]
Solve for [tex]v_i[/tex] and [tex]\theta_i[/tex] :
[tex]\dfrac{3{v_i}^2 \sin^2(\theta_i)}g = \dfrac{2{v_i}^2 \cos(\theta_i) \sin(\theta_i)}g \\\\ \implies 3 \sin^2(\theta_i) = 2 \cos(\theta_i) \sin(\theta_i) \\\\ \sin(\theta_i) (3\sin(\theta_i) - 2 \cos(\theta_i)) = 0[/tex]
Since [tex]0^\circ<\theta_i<90^\circ[/tex], we cannot have [tex]\sin(\theta_i)=0[/tex], so we're left with (e)
[tex]3 \sin(\theta_i) - 2\cos(\theta_i) = 0 \\\\ \implies 3 \sin(\theta_i) = 2\cos(\theta_i) \\\\ \implies \tan(\theta_i) = \dfrac23 \\\\ \implies \boxed{\theta_i = \tan^{-1}\left(\dfrac23\right) \approx 33.7^\circ}[/tex]
Now,
[tex]\cos\left(\tan^{-1}\left(\dfrac23\right)\right) = \dfrac3{\sqrt{13}}[/tex]
[tex]\sin\left(\tan^{-1}\left(\dfrac23\right)\right) = \dfrac2{\sqrt{13}}[/tex]
so it follows that (d)
[tex]R = \dfrac{2{v_i}^2 \times\frac3{\sqrt{13}} \times \frac2{\sqrt{13}}}g \\\\ \implies {v_i}^2 = \dfrac{13Rg}{12} \\\\ \implies \boxed{v_i = \sqrt{\dfrac{13Rg}{12}}}[/tex]
Knowing the initial speed and angle, the initial vertical component of velocity is (c)
[tex]v_y = \sqrt{\dfrac{13Rg}{12}} \sin\left(\tan^{-1}\left(\dfrac23\right)\right) \\\\ \implies v_y = \sqrt{\dfrac{13Rg}{12}} \times \dfrac2{\sqrt{13}} \\\\ \implies \boxed{v_y = \sqrt{\dfrac{Rg}3}}[/tex]
We mentioned earlier that the vertical velocity is zero at maximum height, so the speed of the ball is entirely determined by the horizontal component. (b)
[tex]v_x = \sqrt{\dfrac{13Rg}{12}} \times \dfrac3{\sqrt{13}} \\\\ \implies v_x = \dfrac{\sqrt{3Rg}}{2}[/tex]
Then with [tex]v_y=0[/tex], the ball's speed [tex]v[/tex] is
[tex]v = \sqrt{{v_x}^2 + {v_y}^2} \\\\ \implies v = v_x \\\\ \implies \boxed{v = \dfrac{\sqrt{3Rg}}2}[/tex]
Finally, in the work leading up to part (e), we showed the time to maximum height is
[tex]t = \dfrac{v_i \sin(\theta_i)}g[/tex]
but this is just half the total time the ball spends in the air. The total airtime is then
[tex]2t = \dfrac{2 \times \sqrt{\frac{13Rg}{12}} \times \frac2{\sqrt{13}}}g \\\\ \implies 2t = 2\sqrt{\dfrac R{3g}}[/tex]
and the ball is in the air over the interval (a)
[tex]\boxed{0 < t < 2\sqrt{\frac R{3g}}}[/tex]
