A ball is shot from the ground straight up into the air with initial velocity of 40 ft/sec. Assuming that the air resistance can be ignored, how high does it go

Answers

Answer 1

The height to which the ball attained, given the data is 7.58 m

Data obtained from the questionInitial velocity (u) = 40 ft/s = 40 × 0.3048 = 12.192 m/sFinal velocity (v) = 0 m/s (at maximum height) Acceleration due to gravity (g) = 9.8 m/s²Maximum height (h) =?

How to determine the maximum height

v² = u² – 2gh (since the ball is going against gravity)

0² = 12.192² – (2 × 9.8 × h)

0 = 12.192² – 19.6h

Collect like terms

0 – 12.192² = –19.6h

–12.192² = –19.6h

Divide both side by –19.6

h = –12.192² / –19.6

h = 7.58 m

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Related Questions

Particles q1, 92, and q3 are in a straight line.
Particles q1 = -5.00 x 10-6 C,q2 = -5.00 x 10-6 C,
and q3 = -5.00 x 10-6 C. Particles q₁ and q2 are
separated by 0.500 m. Particles q2 and q3 are
separated by 0.500 m. What is net force on 93?
Remember: Negative forces (-F) will point Left
Positive forces (+F) will point Right
-5.00 x 10-6 C
91
0.500 m-
-5.00 x 10-6 C
92
-5.00 x 10-6 C
93
0.500 m-

Answers

The answer to the question is 1.125  Newton

Formula for electrostatic force is F = ( K q1 q2 )/ r²

where q1 and q2 represent the charges and r represents the distance between them and the value of K is 9 × 10⁹.

Total net force on q3 will be the summation of electrostatic force between q1 and q3 and electrostatic force between q2 and q3, as all the three charges are of same sign and lie in the same line.

Electrostatic force between q1 and q3

r will be 0.500 + 0.500 = 1 m

F = ( K q1 q3 )/ r²

F = ( (9 × 10⁹ ) x (5.00 x 10⁻⁶) x (5.00 x 10⁻⁶) )/ 1²

F₁₃ = 2.25  × 10⁻¹ N

Electrostatic force between q2 and q3

r will be 0.500 m

F = ( K q1 q3 )/ r²

F = ( (9 × 10⁹ ) x (5.00 x 10⁻⁶) x (5.00 x 10⁻⁶) )/ 0.5²

F = (225 × 10⁻³) / (25 × 10⁻²)

F₂₃ = 9 × 10⁻¹ N

Total force on q3 will be F₁₃ + F₂₃

Total force on q3 =  ( 2.25  × 10⁻¹ ) + (9 × 10⁻¹ ) N

Total force on q3 =  ( 2.25  × 10⁻¹ ) + (9 × 10⁻¹ ) N

Total force on q3 =  ( 11.25  × 10⁻¹ ) N

Total force on q3 =  1.125  N

Thus after solving we got the net force on q3 as 1.125  Newton

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