Physics help???

Question

A 0.5 kg basketball is rolling by you at 3.6 m/s. As it goes by, you give it a quick kick perpendicular to its path. Your foot is in contact with the ball for 0.002 s. The ball eventually rolls at an angle of theta = 22 degrees from its original direction.

Determine the magnitude of the average force you applied to the ball.

Any help/explanation would be greatly appreciated!

Answer 1

Suppose the ball gained velocity

= v in the perpendicular direction

=> v cos 22° = 3.6 sin 22°

=> v = 3.6 tan 22° = 1.454 m/s

Average force

= rate of change of momentum

= (0.5) * (1.454) / (0.002) N

= 363.5 N.

Answer 2

The surface is required, for is it a plane surface on which a round objects (sphere) travels by response, if it is cliveted for 0.002, the ball will respond to the injunct, @ theta=-(0)@ erg (0.5 + psi = v*f/2  and stop,  performing this only works if you use a one of your feet, note a foot is an un-covered foot (1 feet)