Answer:
Only the car transforms electrical energy into more than one form of energy.
Explanation:
The motion of the car is mechanical energy but it can also transform into another energy witch is electrical energy
Renee looks out a window. The window is clear, or transparent. This means most of the light that hits the window is:
А
scattered
B
reflected
с
transmitted
D
absorbed
Answer:
the answer is transmitted
Explanation:
6.
ribbon
AA
SON
120 N
Two teams of students are competing in a tug-o-war contest, as shown in the
picture above. How does the ribbon move?
Answer:
The ribbon will move to the right.
Explanation:
To know the the correct answer to the question, we shall determine the net force and direction. This can be obtained as follow:
Force to the right (Fᵣ) = 120 N
Force to the left (Fₗ) = 80 N
Net force (Fₙ) =?
Fₙ = Fᵣ – Fₗ
Fₙ = 120 – 80
Fₙ = 40 N to the right.
From the calculation made above, the net force is 40 N to the right. Thus, the ribbon will move to the right.
What school did Ronald McNair go to and what kind of science did he work in
Answer:
McNair graduated as valedictorian of Carver High School in 1967. In 1971, he received a Bachelor of Science degree in engineering physics, magna cu.m laude, from the North Carolina Agricultural and Technical State University in Greensboro, North Carolina.
how is friction involved in the movement of space
Answer:
Friction can stop or slow down the motion of an object.
Explanation:
The slowing force of friction always acts in the direction opposite to the force causing the motion.
How high above the ground would a 2 kg object need to be in order to have 180 J
of gravitational potential energy?
Answer:
energy= MGH
2*9.8*h=180
h=180/19.6
h=9.32 m
The height of the object above the ground would be equal to 9.18 m.
What is gravitational potential energy?When an object of mass (m) is moved from infinity to a certain point inside the gravitational influence, the amount of work done in displacing it is stored in the form of potential energy and is known as gravitational potential energy.
The mathematical equation for gravitational potential energy can be written as:
Gravitational potential energy = m⋅g⋅h
Where m is the mass, g is the gravitational acceleration and h is the height above the ground.
Given, the mass of the given object, m = 2 Kg
The gravitational potential energy = 180 J
[tex]GPE = m\times g\times h[/tex]
180 = 2 × 9.8 × h
h = 9.18 m
Therefore, the object should be at a height of 9.18 meters in order to have 180 J of gravitational potential energy.
Learn more about gravitational potential energy, here:
https://brainly.com/question/3884855
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8) A train enters a curved horizontal section of the track at a speed of 100 km/h and slows down with constant deceleration to 50 km/h in 12 seconds. If the total horizontal acceleration of the train is 2 m/s2 when the train is 6 seconds into the curve, calculate the radius of curvature of the track for this instant.
Answer:
the radius of curvature of the track for this instant is 266 m
Explanation:
Given that;
The Initial Velocity u = 100 km/h = 100 × [tex]\frac{5}{18}[/tex] = 27.77 m/s
velocity of the train at t=12 s is;
[tex]V_{t=12}[/tex] = 50 km/h = 50 × [tex]\frac{5}{18}[/tex] = 13.89 m/s
now, we calculate the deceleration of the train
[tex]V_{t=12}[/tex] = u + at
13.89 = 27.77 + [tex]a_{t}[/tex]12
[tex]a_{t}[/tex] = (13.89 - 27.77) / 12
[tex]a_{t}[/tex] = -13.88 / 12
[tex]a_{t}[/tex] = - 1.1566 m/s²
Now, the velocity of the train at 6 seconds is;
[tex]V_{t=6}[/tex] = u + at
[tex]V_{t=6}[/tex] = 27.77 + ( - 1.1566 m/s²)6
[tex]V_{t=6}[/tex] = 27.77 - 6.9396
[tex]V_{t=6}[/tex] = 20.83 m/s
The acceleration at t=6 s is;
a = √[ ([tex]a_{t}[/tex] )² + ([tex]a_{n}[/tex])²]
a = √[ ([tex]a_{t}[/tex] )² + ([tex]a_{n}[/tex])²]
we substitute
2m/s² = √[ (- 1.15 )² + ([tex]a_{n}[/tex])²]
4 = (- 1.1566 )² + ([tex]a_{n}[/tex])²
4 = 1.3377 + ([tex]a_{n}[/tex])²
([tex]a_{n}[/tex])² = 4 - 1.3377
([tex]a_{n}[/tex])² = 2.6623
[tex]a_{n}[/tex] = √2.6623
[tex]a_{n}[/tex] = 1.6316 m/s²
Now the radius of curve is;
a = V² / p
[tex]p_{t=6}[/tex] = ( [tex]V_{t=6}[/tex] )² / [tex]a_{n}[/tex]
[tex]p_{t=6}[/tex] = ( 20.83 m/s )² / 1.6316 m/s²
[tex]p_{t=6}[/tex] = 433.8889 / 1.6316
[tex]p_{t=6}[/tex] = 265.9 m ≈ 266 m
Therefore; the radius of curvature of the track for this instant is 266 m
When the material in the mantle cools off near the surface then sinks
down towards the core and get heated again and rises back towards the
surface it is called?
Condensation
О
The water cycle
O
Convection currents
О
Apples and Bananas.
PLEASE HELP THIS IS URGENT ITS FOR A TEST
Answer:
convection currents
Explanation:
Violet pulls a rake horizontally on a frictionless driveway with a net force of 2.0 N for 5.0 m.
How much kinetic energy does the rake gain?
Answer:
10 J.
Explanation:
Given that,
Net force acting on the rake, F = 2 N
Distance moved by the rake, d = 5 m
We need to find the kinetic energy gained by the rake. We know that,
Kinetic energy = work done
So,
K = F×d
K = 2 N × 5 m
K = 10 J
So, 10 J of kinetic energy is gained by the rake.
Violet pulls a rake horizontally on a frictionless driveway with a net force of 2.0 N for 5.0 m.
How much kinetic energy does the rake gain?
Answer: 10 J
Based on your average reaction time, how much time would it take to react to a traffic situation and stop a car traveling at 60 mph (1 mph equals 0.45 m/s) if you could decelerate the car at a rate of -3.4m/s2?
What distance would you travel (in meters) as the car came to a stop in the above situation?
Avg Reaction time: 0.218 ms
Answer:
d = 106.41 m
Explanation:
Given that,
Initial speed of the car, u = 60 mph = 26.9 m/s
The deceleration in the car, a = -3.4 m/s²
The average reaction time, t = 0.218 m/s
It finally stops, final velocity, v = 0
We need to find the distance covered by the car as it come to a stop.
Using third equation of motion to find.
[tex]v^2-u^2=2ad\\\\d=\dfrac{v^2-u^2}{2a}\\\\d=\dfrac{0^2-26.9^2}{2\times (-3.4)}\\\\d=106.41 m[/tex]
So, the car will cover 106.41 m as it comes to a stop.
a point charge q1 = 2.40 uC is held stationary at the origin. A second point charge q2 = -4.30uC moves from the point x= .150 m, y= 0.0 m, to the point x = .250 m, y= 0.0m
a) what is the charge in potential energy of the pair of charges?
b) How much work is done by the electric force on q2
Answer:150M
Explanation:
Which of the following is NOT a characteristic of noble gases?
unreactive
odorless
solid at room temperature
colorless
A constant torque of 3 Nm is applied to an unloaded motor at rest at time t = 0. The motor reaches a speed of 1,393 rpm in 4 s. Assuming the damping to be negligible, calculate the motor inertia in Nm·s2.
Answer:
The moment of inertia of the motor is 0.0823 Newton-meter-square seconds.
Explanation:
From Newton's Laws of Motion and Principle of Motion of D'Alembert, the net torque of a system ([tex]\tau[/tex]), measured in Newton-meters, is:
[tex]\tau = I\cdot \alpha[/tex] (1)
Where:
[tex]I[/tex] - Moment of inertia, measured in Newton-meter-square seconds.
[tex]\alpha[/tex] - Angular acceleration, measured in radians per square second.
If motor have an uniform acceleration, then we can calculate acceleration by this formula:
[tex]\alpha = \frac{\omega - \omega_{o}}{t}[/tex] (2)
Where:
[tex]\omega_{o}[/tex] - Initial angular speed, measured in radians per second.
[tex]\omega[/tex] - Final angular speed, measured in radians per second.
[tex]t[/tex] - Time, measured in seconds.
If we know that [tex]\tau = 3\,N\cdot m[/tex], [tex]\omega_{o} = 0\,\frac{rad}{s }[/tex], [tex]\omega = 145.875\,\frac{rad}{s}[/tex] and [tex]t = 4\,s[/tex], then the moment of inertia of the motor is:
[tex]\alpha = \frac{145.875\,\frac{rad}{s}-0\,\frac{rad}{s}}{4\,s}[/tex]
[tex]\alpha = 36.469\,\frac{rad}{s^{2}}[/tex]
[tex]I = \frac{\tau}{\alpha}[/tex]
[tex]I = \frac{3\,N\cdot m}{36.469\,\frac{rad}{s^{2}} }[/tex]
[tex]I = 0.0823\,N\cdot m\cdot s^{2}[/tex]
The moment of inertia of the motor is 0.0823 Newton-meter-square seconds.
3. Do you think Lynn’s (the protagonist)actions were justifiable by her motives? Why or why not? Please help me Bad Genius the movie
Answer:
I do believe her actions were justified.
Explanation:
Due to the school charging extra fee from her father who makes a modest amount as a teacher. There was sum of money involved that could change how he lived and her.
I do not believe her actions where justified
She had a lot going for her. She could have skipped the hardship of helping grace and pass. She could have easily have gotten a good job with a degree and paid back all the debts owed. Alot of troubles could have been avoided just by doing her own thing.
Concept Simulation 2.3 offers a useful review of the concepts central to this problem. An astronaut on a distant planet wants to determine its acceleration due to gravity. The astronaut throws a rock straight up with a velocity of 17.4 m/s and measures a time of 12.4 s before the rock returns to his hand. What is the acceleration (magnitude and direction) due to gravity on this planet
Answer:
1.40 m/s^2
Explanation:
Given data
Velocity= 17.4 m/s
time= 12.4 seconds
We want to find the acceleration of the rock
We know that
acceleration = velocity/time
Substitute
acceleration= 17.4/12.4
acceleration=1.40 m/s^2
Hence the acceleration is 1.40 m/s^2
Which cell line is pointing to the body?
Answer:
The answer is B .........number 2
Explanation:
An 80 N rightward force is applied to a 10 kg object to accelerate it to the right.
The object encounters a friction force of 50 N.
net force = 30 N
mass = 8.16 kg
acceleration = 3.68 m/s²
Further explanationGiven
80 N force applied
mass of object = 10 kg
Friction force = 50 N
Required
Net force
mass
acceleration
Solution
net forceNet force = force applied(to the right) - friction force(to the left)
Net force = 80 - 50 = 30 N
massGravitational force(downward) : F = mg
m = F : g
m = 80 : 9.8
m = 8.16 kg
accelerationa = F net / m
a = 30 / 8.16
a = 3.68 m/s²
A skydiver is using wind to land on a target that is 120 m away horizontally. The skydiver starts from a height of 70 m and is falling vertically at a constant velocity of 7.0 m/s downward with their parachute open (terminal velocity). A horizontal gust of wind helps push them towards the target. What must be their total speed if they want to just hit their target
Answer:
13.9 m/s.
Explanation:
Since the vertical velocity of the skydiver is constant at v = 7.0 m/s, we find the time, t it takes him to drop from a height of h = 70 m.
So, distance = velocity time
h = vt
t = h/v = 70 m/7 m/s = 10 s
This is also the time it takes him to move horizontally a distance of d = 120 m to the target.
So, his horizontal velocity is v' = distance/time = d/t = 120m/10 s = 12 m/s.
Since both vertical and horizontal velocities are perpendicular, we add them vectorially to obtain the skydivers total speed, V.
So, V = √(v² + v'²)
= √((7.0 m/s)² + (12.0 m/s)'²)
= √(49 m²/s² + 144 m²/s²)
= √(193 m²/s²)
= 13.9 m/s.
The direction of this velocity is Ф = tan⁻¹(v/v')
= tan⁻¹(7 m/s/12 m/s)
= tan⁻¹(0.5833)
= 30.3°
You are on the Pirates of the Caribbean attraction in the Magic Kingdom at Disney World. Your boat rides through a pirate battle, in which cannons on a ship and in a fort are firing at each other. While you are aware that the splashes in the water do not represent actual cannonballs, you begin to wonder about such battles in the days of the pirates. Sup-pose the fort and the ship are separated by 75.0 m. You see that the cannons in the fort are aimed so that their cannon-balls would be fired horizontally from a height of 7.00 m above the water.
(a) You wonder at what speed they must be fired in order to hit the ship before falling in the water.
(b) Then, you think about the sludge that must build up inside the barrel of a cannon. This sludge should slow down the cannonballs. A question occurs in your mind: if the can-nonballs can be fired at only 50.0% of the speed found ear-lier, is it possible to fire them upward at some angle to the horizontal so that they would reach the ship?
Answer:
a) v₀ₓ = 62.76 m / s, b) θ₁ = 17.6º, θ₂ = 67.0º
Explanation:
We can solve this exercise using the projectile launch ratios
a) Let's find the time it takes for the bullet to reach the water level
y = y₀ + v_{oy} t - ½ g t²
when it reaches the water its height is zero y = 0, as the bullet is fired horizontally its initial vertical velocity is zero
0 = y₀ + 0 - ½ g t²
t =[tex]\sqrt{2y_o/g}[/tex]
t = [tex]\sqrt{2 \ 7 /9.8}[/tex]
t = 1,195 s
now we can calculate the speed with the horizontal movement
x = v₀ₓ t
v₀ₓ = x / t
v₀ₓ = 75.0 / 1.195
v₀ₓ = 62.76 m / s
b) if the speed of the bullets is half of that found
v₀ = 62.76 / 2 = 31.38 m / s
let's write the expressions for the distance
x = v₀ cos θ t
y = y₀ + v_{oy} sin θ t - ½ g t²
t = [tex]\frac{x}{v_o \ cos \theta}[/tex]
we substitute
[tex]0 = y_o + v_o sin \theta \ \frac{x}{v_o \cos \thetay} - 1/2 g \ (\frac{x}{v_o \ cos \theta})^2[/tex]
[tex]0 = y_o + x tan \theta - \frac{1}{2} g \ \frac{x^2}{ v_o^2 \ cos^2 \theta}[/tex]
let's use the identified trigonometry
sec² θ = 1 + tan² θ
sec θ = 1 / cos θ
we substitute
[tex]0 = y_o + x tan \theta - \frac{g x^2}{2 v_o^2} ( 1 + tan^2 \theta)[/tex]
[tex]\frac{g x^2}{2v_o^2} tan^2 \theta - x tan \theta + \frac{gx^2}{2v_o^2} - y_o = 0[/tex]
we change variable
tan θ = H
[tex]\frac{gx^2}{2 v_o^2 } H^2 - x H + \frac{gx^2}{2v_o^2}-y_o =0[/tex]
we subtitle the values
[tex]\frac{9.8 \ 75^2}{2 \ 31.38^2} H^2 - 75 H + \frac{9.8 \ 75^2}{2 \ 31.38^2}-7 =0[/tex]
27.99 H² - 75 H + 20.99 = 0
H² - 2.679 H + 0.75 = 0
we solve the quadratic equation
H = [2.679 ± [tex]\sqrt{2.679^2 - 4 0.75}[/tex]] / 2
H = [2,679 ± 2,044] / 2
H₁ = 0.3175
H₂ = 2.3615
now we can find the angles
H₁ = tan θ₁
θ₁ = tan⁻¹ H₁
θ₁ = tan⁻¹ 0.3175
θ₁ = 17.6º
θ₂ = 67.0º
for these two angles the bullet hits the boat
In traveling a distance of 2.3 km between points A and D, a car is driven at 83 km/h from A to B for t seconds and 41 km/h from C to D also for t seconds. If the brakes are applied for 4.4 seconds between B and C to give the car a uniform deceleration, calculate t and the distance s between A and B.
Answer:
- time t taken for car to travel is 64.57 s
- distance travelled between A and B is 1.4887 km
Explanation:
Given the data in the question;
[tex]U_{BC}[/tex] = 83 km/h = ( 83×1000 / 60×60) = 23.0555 m/s
[tex]U_{CD}[/tex] = 41 km/h = ( 41×1000 / 60×60) = 11.3888 m/s
now, we calculate the acceleration;
a = ( [tex]U_{BC}[/tex] - [tex]U_{CD}[/tex] ) / t
we substitute
a = ( 23.0555 - 11.3888 ) / 4.4
a = 11.6667 / 4.4
a = 2.6515 m/s²
Now equation for displacement from BC
[tex]S_{BC}[/tex] = [tex]U_{BC}[/tex]t + 1/2.at²
we substitute
[tex]S_{BC}[/tex] = 23.0555×4.4 + 1/2×a×(4.4)²
we substitute -2.6515m/s² for a
[tex]S_{BC}[/tex] = 23.0555×4.4 + 1/2×(-2.6515)×(4.4)²
= 101.4442 - 25.6665
[tex]S_{BC}[/tex] = 75.7792 m
Now, for total distance covered = 2.3km = ( 2.3×1000) = 2300m
so
[tex]S_{AB}[/tex] + [tex]S_{BC}[/tex] + [tex]S_{CD}[/tex] = 2300 m
we substitute substitute
[tex]S_{AB}[/tex] + 75.7792 m + [tex]S_{CD}[/tex] = 2300 m
[tex]S_{AB}[/tex] + [tex]S_{CD}[/tex] = 2300 - 75.7792
[tex]S_{AB}[/tex] + [tex]S_{CD}[/tex] = 2224.2208 m
so we substitute 23.0555t for [tex]S_{AB}[/tex] and 11.3888t for [tex]S_{CD}[/tex]
23.0555t + 11.3888t = 2224.2208
34.4443t = 2224.2208
t = 2224.2208 / 34.4443
t = 64.57 s
Therefore, time t taken for car to travel is 64.57 s
Distance Between A to B
[tex]S_{AB}[/tex] = t × [tex]U_{AB}[/tex]
we substitute
[tex]S_{AB}[/tex] = 64.57 s × 23.0555
[tex]S_{AB}[/tex] = 1488.69 m
[tex]S_{AB}[/tex] = 1.4887 km
Therefore, distance travelled between A and B is 1.4887 km
true or false solubility can be used to identify an unknown substance
What is the displacement of the particle in the time interval 7 seconds to 8 seconds?
Answer:
it 1.5 meters
Explanation:
if u could put the option number it will be cool and hope it help and if it doesnt am really sorry ;)
Which of the following is an example of Newton's Third Law?* O A stack of pennies will not move unless you flick them over. O Falling off of a skateboard after you run into a curb A ball hits the ground and the ground pushes up on it with the same force
Answer:
A ball hits the ground and the ground pushes up on it
Explanation:
Newton's third law basically states that for every action, there's a reaction.
a ball hitting the ground would be the action. the ground pushing up on it with the same force is the reaction.
Hope this Helps!!! :)
What makes electromagnets useful for sorting metals in recycling centers?
O A. The current can be turned on to pick up items containing all
metals and turned off to drop them.
O B. The current can be turned off to pick up items containing all
metals and turned off to drop them.
O C. The current can be turned on to pick up items containing iron and
turned off to drop them.
D. The current can be turned off to pick up items containing iron and
turned on to drop them.
C
It is right because I took this and I got this answer correct
The head of a rattlesnake can accelerate at 50 m/s2 in striking a victim. If a car could do as well, how long would it take to reach a speed of 100 km/h from rest
Answer:
the time for the car to reach the final velocity is 0.56 s.
Explanation:
Given;
acceleration of the car, a = 50 m/s²
final velocity of the car, v = 100 km/h = 27.778 m/s
the initial velocity of the car, u = 0
The time for the car to reach the final velocity is calculated as;
v = u + at
27.778 = 0 + 50t
27.778 = 50t
t = 27.778 / 50
t = 0.56 s
Therefore, the time for the car to reach the final velocity is 0.56 s.
15 points!!:-) Need help ASAP!
When does a compass NOT point towards magnetic north?
A.during a solar eclipse, which changes
Earth's magnetic field.
B. When there is another magnet close by.
C.When there is an unused battery close by.
D. When there is a coil of copper wire close by.
Answer:
b i think so because it makes senes
Answer: When there is another magnet close by.
Explanation: The needle of a compass is itself a magnet, and thus the north pole of the magnet always points north, except when it is near a strong magnet. ... When you take the compass away from the bar magnet, it again points north. So, we can conclude that the north end of a compass is attracted to the south end of a magnet.
Potential energy is defined as
A. energy of motion
B. moving another object
C. stored energy
Answer:
c
Explanation:
it is stored energy because it is built up in said object
the water behind hoover dam in nevada is 206 m higher than the colorado river below it. at what rate must water pass through the hydraulic turbines of this dam to produce 100 mw of power if the turbines are 100 percent efficient
Answer:
the required mass flow rate is 49484.37 kg/s
Explanation:
Given the data in the question;
we first determine the relation for mass flow rate of water that passes through the turbine;
so the relation for net work on the turbine due to the change in potential energy considering 100% efficiency is;
[tex]W_{net}[/tex] = m ( Δ P.E )
so we substitute (gh) for ( Δ P.E );
[tex]W_{net}[/tex] = m (gh)
m = [tex]W_{net}[/tex] / gh
so we substitute our given values into the equation
m = 100 MW / ( 9.81 m/s²) × 206 m
m = ( 100 MW × 10⁶W/MW) / ( 9.81 m/s²) × 206 m
m = 10 × 10⁷ / 2020.86
m = 49484.37 kg/s
Therefore, the required mass flow rate is 49484.37 kg/s
(a) The electric potential due to a point charge is given by V = kq⁄r where q is the charge, r is the distance from q and k = 8.99 × 109 ????m2 C 2 ⁄ . Show, in detail, that the SI unit of electric potential is a volt. ( b ) What are the equipotential lines? (c) How are equipotential lines used to obtain the electric field lines? (
Answer:
a) [volts] = [N m / C],
b) The lines or surface that has the same potential are called equipotential
c) the equipotential lines must also be perpendicular to the electric field lines
Explanation:
a) find the units of the volt
the electric potential energy is
V = k q / r
V = [N m² / C²] C / m
V = [N m / C]
The electric potential is defined as
V = E .s
V = [N / C] [m]
V = [N m / C] = [volt]
we see that in the two expressions the same result is obtained therefore the volt is
[volts] = [N m / C]
b) The lines or surface that has the same potential are called equipotential surfaces, the great utility of these lines or surfaces is that a face can be displaced on it without doing work.
c) The electric potential is defined as the gradient of the electric field
v = [tex]- \frac{dE}{dx} i^[/tex]
therefore the equipotential lines must also be perpendicular to the electric field lines
(a) The unit of electric potential is volts.
(b) Equipotential lines are lines along which the electric potential is constant.
(c) Equipotential lines are obtained by drawing a line perpendicular to electric field lines.
SI unit of electric potentialElectric potential is the work done in moving a unit positive from infinite to a point in the electric filed.
Electric potential = EQ
Electric potential = (V/C) x (C)
Electric potential = Volts.
Equipotential linesThis is a line along which the electric potential is constant.
How to obtain equipotential linesEquipotential lines are obtained by drawing a line perpendicular to electric field lines.
Learn more about equipotential lines here: https://brainly.com/question/12074544
The deepest part of the ocean is the Challenger Deep, at 10,900 m. The depth was first measured in 1875 by the HMS Challenger by depth sounding (which does not involve sound waves). If you were to measure the depth by echo sounding (which does involve sound), what would you expect the time for a sound pulse at the surface to return in s, naively assuming a constant sound velocity throughout the ocean
Answer:
t = 14.53 s
Explanation:
The speed of a wave is constant and is given by
v = [tex]\sqrt{ \frac{B}{ \rho} }[/tex]
in this exercise they indicate that we assume the constant velocity, therefore we can use the uniform motion relations
v = x / t
t = x / v
in this case the sound pulse leaves the ship and must return so the distance is
x = 2d
where d is the ocean depth d = 10900m and the speed of sound in seawater is v = 1500 m / s
let's calculate
t = 2 10900/1500
t = 14.53 s
Could I get help on this question please
Answer:
124.51 m
Explanation:
From the question given above, the following data were obtained:
Initial velocity (u) = 49.4 m/s
Final velocity (v) = 0 m/s (at maximum height)
Maximum height (h) =?
NOTE: Acceleration due to gravity (g) = 9.8 m/s²
The maximum height to which the cannon ball attained before falling back can be obtained as illustrated below:
v² = u² – 2gh ( since the ball is going against gravity)
0² = 49.4² – (2 × 9.8 × h)
0 = 2440.36 – 19.6h
Collect like terms
0 – 2440.36 = –19.6h
–2440.36 = –19.6h
Divide both side by –19.6
h = –2440.36 / –19.6
h = 124.51 m
Therefore, maximum height to which the cannon ball attained before falling back is 124.51 m
