A tennis ball on Mars, where the acceleration due to gravity is 0.379 of a g and

Question

A tennis ball on Mars, where the acceleration due to gravity is 0.379 of a g and air resistance is negligible, is hit directly upward and returns to the same level 7.50 s later.
A. How high above it’s original point did the ball go? (meters)
B. How fast was it moving just after being hit? (m/s) A tennis ball on Mars, where the acceleration due to gravity is 0.379 of a g and air resistance is negligible, is hit directly upward and returns to the same level 7.50 s later.
A. How high above it’s original point did the ball go? (meters)
B. How fast was it moving just after being hit? (m/s)
A. How high above it’s original point did the ball go? (meters)
B. How fast was it moving just after being hit? (m/s)

Explanation / Answer

Using kinematic equation,

V = u + at

0 = u + at

Time, t = - u/a

Given, 2t = 7.5

=> 2u/a = 7.5

First we will solve part B

B)

Initial speed, u = 7.5 a/2 = 7.5 x 0.379g/2 = 13.93 m/s

A)

V^2 - u^2 = 2as

0 - 13.93^2 = 2 x 0.379g x s

Distance, s = (13.93)^2/(2 x 0.379 x 9.8) = 26.115 m

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