Question

A punter drops a ball from rest vertically 1 meter down onto his foot. The ball leaves the foot with a speed of 25 m/s at an angle 42° above the horizontal. What is the impulse delivered by the foot? (Use 0.41 kg for the mass of the ball. Assume the punter faces the +x-direction and that the +y-direction is upward.)

Find the impulse in kg · m/s and the direction in degrees counterclockwise from the +x-axis.

Answer 1

Velocity of ball just before hitting the foot = root( 2 g h) along -ve y axis ( h is height from where ball is dropped)
V(i) = - root(2*9.8*1)j = -4.43 j
Velociyt of ball, after it leaves the foot, V(f) = 25 cos 42 i + 25 sin42j
                                                          = 18.6i + 16.73j
change in velocity = V(f) - V(i) = 18.58i + 21.16j
Impulse = change in momentum = mass* change in velocity
                                                 = 7.62i + 8.67j
magnitude of change in momentum = root(7.62² + 8.67²) = 11.55 kg m/sec
angle with +ve axis in counterclockwise, Theta = tan^-1(8.67/7.62) = 48.7 deg