Question

A soccer player kicks a 0.43 kg soccer ball down a smooth hill 18 m high with an initial speed of 7.4 m/s.

a) Calculate the ball’s speed as it reaches the bottom of the hill.

b) The soccer player stands at the same point on the hill and gives the ball a kick up the hill at 4.2 m/s. The ball moves up the hill, comes to rest, and rolls back down the hill. Determine the ball’s speed as it reaches the bottom of the hill.

Answer 1

Given mass of ball =0.43kg

Height=18m

Initial velocity =7.4 m/s

Now

A.

initial KE = 1/2(mv^2) = 11.7734 Joules.

And KE at bottom of hill = (0.43 x g x 18) + 11.7734

=87.6254 Joules. (g = 9.8 used).

Speed at bottom = sqrt(2KE/m)= 20.19m/sec.

So ball velocity willl be 4.2m/sec when it passes the player.

B.

KE=1/2 (mv^2) = 3.7926 Joules.

KE at bottom = (0.43 x 9.8 x 18) + 3.7926 = 79.6446 Joules.

V at bottom = 19.25m/sec.

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