Question

In a football game, a 96 kg receiver leaps straight up in the air to catch the 0.42 kg ball the quarterback threw to him at a vigorous 17 m/s, catching the ball at the highest point in his jump.

Answer 1

Answer:

The speed of the receiver is 0.074 m/s.

Explanation:

Given that,

Mass of the receiver, m = 96 kg

Mass of the ball, m' = 0.42 m/s

Initially the receiver is at rest, u = 0

Initial speed of the ball, u' = 17 m/s

We need to find the speed of the receiver after the collision. It is a case of conservation of momentum. Let V is the speed of the receiver.

$mu+m'u'=(m+m')V$

$m'u'=(m+m')V$

$V=\dfrac{m'u'}{m+m'}$

$V=\dfrac{0.42\times 17}{96+0.42}$

V = 0.074 m/s

So, the speed of the receiver is 0.074 m/s. Hence, this is the required solution.