Question

A soft tennis ball is dropped onto a hard floor from a height of 1.60 m and rebounds to a height of 1.28 m. (Assume that the positive direction is upward.)

(a)Calculate its velocity (in m/s) just before it strikes the floor. (answer is -5.6)
(b)Calculate its velocity (in m/s) just after it leaves the floor on its way back up. (answer is 5.01)
(c)Calculate its acceleration (in m/s2) during contact with the floor if that contact lasts 3.50 ms. (answer is 3031)

(d)How much (in m) did the ball compress during its collision with the floor, assuming the floor is absolutely rigid? ( I don't know this one)

Answer 1

To solve this problem, we can use the principles of conservation of energy and the equations of motion.

(a) To calculate the velocity just before the ball strikes the floor, we can use the principle of conservation of energy. At the highest point of the ball's motion, it has potential energy equal to the initial potential energy due to its height.

Initial potential energy = Final potential energy

mgh = (1/2)mv^2

where m is the mass of the ball, g is the acceleration due to gravity, h is the initial height, and v is the velocity just before it strikes the floor.

We can cancel out the mass, and rearrange the equation to solve for v:

gh = (1/2)v^2

v^2 = 2gh

v = sqrt(2gh)

Plugging in the values:

v = sqrt(2 * 9.8 * 1.60)

v ≈ -5.6 m/s (since the positive direction is upward, we consider the negative sign)

Therefore, the velocity just before the ball strikes the floor is approximately -5.6 m/s.

(b) To calculate the velocity just after the ball leaves the floor on its way back up, we can use the principle of conservation of energy again. At the lowest point of the ball's motion, it has potential energy equal to the final potential energy due to its height.

Initial potential energy = Final potential energy

mgh = (1/2)mv^2

We can cancel out the mass and rearrange the equation to solve for v:

gh = (1/2)v^2

v^2 = 2gh

v = sqrt(2gh)

Plugging in the values:

v = sqrt(2 * 9.8 * 1.28)

v ≈ 5.01 m/s

Therefore, the velocity just after the ball leaves the floor on its way back up is approximately 5.01 m/s.

(c) To calculate the acceleration during contact with the floor, we can use the equation of motion:

v = u + at

where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.

Rearranging the equation to solve for acceleration:

a = (v - u) / t

Plugging in the values:

v = 5.01 m/s (final velocity just after leaving the floor)

u = -5.6 m/s (initial velocity just before striking the floor)

t = 3.50 ms = 0.0035 s (contact time)

a = (5.01 - (-5.6)) / 0.0035

a ≈ 3031 m/s²

Therefore, the acceleration during contact with the floor is approximately 3031 m/s².