Abaseball catcher is performing a stunt for a television commercial. he will catch a baseball (mass 145 g) dropped from a height of 66.0 m above his glove. his glove stops the ball in 0.0118 s. what is the force (in n) exerted by his glove on the ball? (indicate the direction with the sign of your answer. assume the baseball is traveling in the negative direction.)

Explanation:

The ball is going down with velocity. It must have momentum . It is stopped by

catcher so that its momentum becomes zero . There is change in momentum . So force is applied on the ball  by the gloves  .

The rate of change of momentum gives the magnitude of force. This force must be in upward direction to stop the ball. So force is in positive direction .

Let us measure the force applied on the ball .

Final velocity after the fall by 66 m

V = √ 2gh

= √ 2x9.8 x 66

35.97 m /s

Momentum = m v

0.145 x 35.97

= 5.2156 kgms⁻¹

Change in momentum

= 5.2156 - 0

= 5.2156

Rate of change of momentum

=  Change of momentum / time = 5.2156 / .0118

= 442 N

if R be the force exerted by gloves to stop the ball

R - mg represents the net force which stops the ball so

R - mg = 442

R = 442 + mg

= 442 + .145 x 908

443.43 N