Question

I drop a tennis ball from a height of h = 5.52 m above an unrealistically huge pile of thick shaving gel. The ball plunges to a depth of d = 18.0 m in the gel before stopping in the gel, at some unknown distance above the floor.

a) What is the magnitude of the ball's acceleration while it's falling through the air?

b) Calculate the ball's impact speed v (its speed just before it hits the gel).

c) Explain why we should expect the ball's acceleration to be smaller while it's moving through the gel instead of the air; that is, why it should speed up (while falling through the air) more rapidly than it slows down (while stopping in the gel). We're comparing magnitudes only, not taking signs into account.

d) Calculate the magnitude a of tennis ball’s stopping acceleration (how rapidly it slowed down in the gel, assuming this occurred at a constant rate).

e) How much time t did it take the ball to stop, once it hit the gel? This isn't the time from release to stopping, or from release to impact, but from impact to stopping only.

Answer 1

While falling through the air, ball's motion is considered to be a free fall motion. Thus, the acceleration it possess would be the acceleration due to gravity. Hence, the magnitude of the ball's acceleration is g= 9.8 m/s? Write the third equation of motion for free fall, v2 = u? +2gh Substitute the value 0 for u, 5.52 m for h and 9.8 m/s2 for g. v2 =(0) +2(9.8 m/s)(5.52 m) v2 =108.192 mºls? v=10.4 m/s Hence, the ball's impact speed v is 10.4 m/s.

Ball's motion in the gel includes another force which is called viscous force. Viscous force always acts in the opposite direction of the motion. Viscosity depends on the density of the medium. Since density of the gel is greater than the density of air, the viscous force would be greater. Hence, the ball's downward acceleration will be smaller while moving through the gel instead of the air. d)

Write the third equation of motion for the motion through the gel, 02 = 12 + 2ad Here, a is the stopping acceleration of the ball in the gel. Thus, a=- Substitute the value 10.4 m/s for v and 18 m for d. (10.4 m/s) a= 2(18 m) = -0.289 m/s2 Here negative sign suggests that the ball has the retardation. Hence, the magnitude of stopping acceleration in gel is 0.289 m/s?

Write the first equation of motion for the ball moving through gel, 0=y+at Here, t is the time taken by the ball to stop. Thus, t=- Substitute the calculated value 10.4 m/s for v and -0.289 m/s2 for a. (10.4 m/s) (-0.289 m/s?) = 35.99 $ Hence, the time t it takes the ball to stop is 35.99 sec