A soft tennis ball is dropped onto a hard floor from a height of 1.75 m and rebo

Question

A soft tennis ball is dropped onto a hard floor from a height of 1.75 m and rebounds to a height of 1.38 m. (Assume that the positive direction is upward.)

(a)

Calculate its velocity (in m/s) just before it strikes the floor.

m/s

(b)

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

m/s

(c)

Calculate its acceleration (in m/s2) during contact with the floor if that contact lasts 3.50 ms.

m/s2

(d)

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

m

Explanation / Answer

a) gh = ½v²
v² = 2gh = 2 * 9.8*·1.75 m²/s²
v = -5.86 m/s . . [downward]

b) gh = ½v²
v² = 2gh = 2* 9.8 *1.38 m²/s²
v = 5.20 m/s . . [upward]

c) a[ave] = v/t = (5.20 - [-5.86]) [m/s] / 0.00350 s = 3160 m/s² . . [upward]

d) Assuming, as we did in part (c), that the upward acceleration is constant, you will need to find what portion, T, of the t = 3.50 ms was spent compressing the tennis ball. Then use
s = ½aT² . . where
s = travel (compression) distance
a = acceleration found in part (c)

Assuming constant acceleration,
T = ( |v| / v) t = |v| / a[ave]
= 5.86·0.00350 s / (5.20 + 5.86) = 1.854 ms

s = ½aT² = 0.00543 m = 5.43 mm

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