Question

A soft tennis ball is dropped onto a hard floor from a height of 1.85 m and rebounds to a height of 1.39 m. (Assume that the positive direction is upward.)
(a) Calculate its velocity just before it strikes the floor.

m/s
(b) Calculate its velocity just after it leaves the floor on its way back up.

m/s
(c) Calculate its acceleration during contact with the floor if that contact lasts 3.50 ms.

m/s2
(d) How much did the ball compress during its collision with the floor, assuming the floor is absolutely rigid?

m

Answer 1

What have you thought of doing, so far? Note that if you drop this ball, its initial velocity is zero and its acceleration is given by gravity, which here is a = -9.8 m/(sec^2)

The equation for velocity would then be\[v=v _{0}+at\]

Answer 2

ok

Answer 3

can you calculate the length of time the dropped ball requires to touch the floor? The appropriate equation of motion, based upon the equation for velocity (above) is\[s=s _{0}+v _{0}t+\frac{ 1 }{ 2 }a t^2.\]

Answer 4

what do the a stand for?

Answer 5

"a" stands for "acceleration." In this case the only acceleration present is that due to gravity, which in turn is \[-9.8\frac{ m }{ \sec^2 }\]

Answer 6

Here the acceleration is treated as a constant.
The equation of motion is all you need to calculate how long it takes for the ball to fall a distance of 1.85 m.

Answer 7

k give me a few seconds

Answer 8

im lost how do you solve for t?

Answer 9

\[s=s _{0}+v _{0}t+\frac{ 1 }{ 2 }a t^2\]

Answer 10

When the ball has actually hit the floor, we say s = 0 meters.
What was the initial height of the center of the ball? (It's given.)
What was the initial velocity of the ball? (It's given.)
What is "a" for this vertical motion? (I've given it to you.)
Substituting these values into the above equation will enable you to solve for t^2, and then to solve for t.

Answer 11

Prefer you show exactly how you got your answer.

Answer 12

1.85+0t+1/2at^2?

Answer 13

You'll need to write an equation (instead of an expression such as yours, above). When the ball hits the ground, it will compress; the floor itself will not. Assume that we set "s" in my distance equation equal to 0. It's the initial distance that was 1.85 m.

This leaves you with 0 = 1.85 m + 0t + (1/2) a t^2. What is the value of "a"? Solve the resulting equation for t^2, please, and then for t.


How far up does the ball travel on its first rebound? That's "distance traveled," represented by s.

Part b involves finding the initial velocity, v_0, of the ball on its first rebound.