Solution Found!
Answer: A tennis ball on Mars, where the acceleration due
Chapter 2, Problem 39E(choose chapter or problem)
A tennis ball on Mars, where the acceleration due to gravity is 0.379 g and air resistance is negligible, is hit directly up-ward and returns to the same level 8.5 s later. (a) How high above its original point did the ball go? (b) How fast was it moving just after it was hit? (c) Sketch graphs of the ball’s vertical position, vertical velocity, and vertical acceleration as functions of time while it’s in the Martian air.
Questions & Answers
QUESTION:
A tennis ball on Mars, where the acceleration due to gravity is 0.379 g and air resistance is negligible, is hit directly up-ward and returns to the same level 8.5 s later. (a) How high above its original point did the ball go? (b) How fast was it moving just after it was hit? (c) Sketch graphs of the ball’s vertical position, vertical velocity, and vertical acceleration as functions of time while it’s in the Martian air.
ANSWER:Step 1 of 4
The acceleration due to gravity on Mars is \({g_{mars}} = 0.379g\). The tennis ball is hit directly upward from a height \({y_0} = 0 \), returning to the same height y = 0 after time t =8.5 s.
The acceleration of the tennis ball is,
\({a_y} = - {g_{mars}}\)
Use the relation to find the velocity just after the hit as,
\(y = {y_0} + {v_{0t}}t + \frac{1}{2}{a_y}{t^2}\)
Substitute the values in the above formula, and we can write,
\(0 = 0 + {v_{0t}}\left( {8.5} \right) + \frac{1}{2}\left( { - 0.379 \times 9.8} \right){8.5^2}\)
\({v_{0t}} = \frac{1}{2}\left( {0.379 \times 9.8} \right)8.5\)
\({v_{0t}} = 15.8\;{\rm{m/s}}\)