the velocity changes from 32.8m/s downward, to 21.2m/s upward in 4.08ms
acceleration is the rate of velocity change
m/s per sec , or m/s^2
put units in your equation
...(21.2m/s - -32.8m/s) / .00408s
...13200m/s/s ... 13200m/s^2
FYI ... watch your significant figures
An electronic line judge camera captures the impact of a
57.0-g
tennis ball traveling at
32.8 m/s
with the side line of a tennis court. The ball rebounds with a speed of
21.2 m/s
and is seen to be in contact with the ground for
4.08 ms.
What is the magnitude of the average acceleration of the ball during the time it is in contact with the ground? Assume one-dimensional motion.
The answer is 13235m/s^2 but I don't understand why the unit is in m/s^2 and not just m? Can somebody explain that? I use the equation (Vf-Vi)/T to sole this question.
W
1 answer