A 1.4 kg ball is dropped from an unknown height, h, and hits the ground with a velocity of 15.0 m/s and rebounds with a velocity of 11.0 m/s. If the ball is in contact with the ground for 1/35 of a second find the average force the ground applies to the ball.

Question

anonymous · Answer

[F =m( (v-u) 	)] where F=force v=final velocity u=initial velocity m=mass t=time. Using this formula impulse we get a answer of -196N

anonymous · Answer

I tried that but it wasn't right D: the answer is suppose to be 1288N, thanks for trying though:)

anonymous · Answer

yes the final velocity should be considered as "-11" now the answer will be 1274 the reason for taking final velocity as negative is that it is in the opposite direction as the initial velocity

anonymous · Answer

average force = change of momentum of the ball
impact = Ns = mv - mu
N =   (mv-mu)/ s
    = (1.4*15 --1.4*11) /  1/35 =  1274 newtons

anonymous · Answer

THANKS SO MUCH GUYS! This really helped!!!