In an American football game, an 80 Kg receiver jumps straight in the air and ca

Question

In an American football game, an 80 Kg receiver jumps straight in the air and catches a 0.5 Kg football thrown towards him. Just before the catch, the receiver is a height 0.5m off of the ground. And the football is traveling horizontally at a speed of 40 m/s (90 mph). Assuming the catch is a perfectly inelastic collision (the ball and the player move together afterwards): how fast are the ball and player moving immediately after the catch? what is the total kinetic energy immediately after the catch? how much energy was lost during collision?

Explanation / Answer

We have to find velocity of football player

Potential energy = kinetic energy

m1gh = (1/2)m1*u1^2

so u1 = sqrt (2gh) = sqrt (2*9.8*0.5)

u1 = 3.13 m/s

Now

1) In perfectly inelastic collision, final velocity is same for both bodies and it is V

So V = (m1u1+m2u2)/(m1+m2)

= ((80*3.13)+(0.5*40))/(80+0.5)

= 270/80.5

= 3.35 m/s

2) Kinetic energy is

KE = (1/2)*(m1+m2)*V^2

= (1/2)*80.5*(3.35)^2

= 451.7 J

3) To find lost in energy, we have to calculate initial KE for player as well as for football

So KE(initial)= ((1/2)m1*u1^2)+((1/2)m2*u2^2)

= (1/2)*80*(3.13)^2)+(1/2)*0.5*(40)^2)

= 391.87+400

= 791.87 J

So lost in energy = KE(initial) - KE(final)

= 791.87 -451.7

= 340.17 J

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