Question

a tennis ball of mass 57.0 g is held just above a basketball of mass 607 g. with their centers vertically aligned, both balls are released from rest at the same time, to fall through a distance of 1.05 m, as shown in the figure below. an illustration shows a tennis ball placed on top of a basketball with a downward-pointing arrow shown to the side. (a) find the magnitude of the downward velocity with which the basketball reaches the ground. use conservation of mechanical energy or the appropriate kinematic equations to find the speed attained by free fall through the given distance. m/s (b) assume that an elastic collision with the ground instantaneously reverses the velocity of the basketball while the tennis ball is still moving down. next, the two balls meet in an elastic collision. to what height does the tennis ball rebound?

Answer 1

The tennis ball rebounds to a height of approximately 20.9 meters.

Here's the solution to the problem:

Let v_tennis_after be the velocity of the tennis ball after the collision and v_basketball_after be the velocity of the basketball after the collision. We can use the conservation of momentum equation to solve for these velocities:

m_tennis * v_tennis_before + m_basketball * v_basketball_before = m_tennis * v_tennis_after + m_basketball * v_basketball_after

where:

m_tennis is the mass of the tennis ball (57.0 g)

v_tennis_before is the velocity of the tennis ball before the collision (4.64 m/s downwards)

m_basketball is the mass of the basketball (607 g)

v_basketball_before is the velocity of the basketball before the collision (4.64 m/s upwards)

Plugging in the values, we get:

57 g * -4.64 m/s + 607 g * 4.64 m/s = 57 g * v_tennis_after + 607 g * v_basketball_after

v_tennis_after = 20.14 m/s upwards

v_basketball_after = 0.84 m/s upwards

The tennis ball rebounds with a velocity of 20.14 m/s upwards, which means it will reach a height of:

h = (v² - v₀²)/(2g) = (20.14 m/s)²/(2 * 9.81 m/s²) ≈ 20.9 meters