What is the speed of the ball traveling? (momentum conservation)

Question

A 90.0 kg man jumps straight up (with no horizontal component of velocity) and catches a horizontally traveling ball. He catches the 0.430 kg ball exactly at the peak of his leap, which is 0.538 meters off the ground. He hits the ground 0.0333 meters away from where he leapt. When it was caught, how fast was the ball travelling?

Answer 1

Let

M = 90 kg be the mass of the man,

m = 0.43 kg be the mass of the ball,

h = 0.538 m be the height of the jump,

d = 0.0333 m be the distance where he lands,

v (to be computed) be the velocity of the ball,

V (unknown) be the horizontal velocity of the man together with the ball,

t (unknown) be the time it took to fall to the ground,

g = 9.81 m/s² be gravitational acceleration.

We have three unknowns and the following three equations

mv = (M+m)V (1)

d = Vt (2)

h = gt²/2 (3)

Equation (1) is conservation of momentum in the horizontal dimension.

The horizontal momentum of the ball before the catch is mv,

the horizontal momentum of the man before the catch is 0,

the horizontal momentum of the man with the ball is (M+m)V.

Equation (2) says that while the man is falling down, he is travelling horizontally with constant velocity V.

Equation (3) is the formula for distance travelled with constant acceleration g.

The solution to the three equations is

t = √(2h/g) from equation (3)

V = d/t = d√(g/2h) from equation (2)

v = (M+m)V/m = (M+m)d√(g/2h)/m from equation (1)

Substituting actual numbers

v = (90 + 0.43)×0.0333×√(9.81/1.076)/0.43 ≈ 21.1 m/s