A3 kg steel ball falls down vertically and strikes a floor with a speed of 12 m/s. it bounces off upward with the same speed. if the ball is in contact with the floor for 0.12 s, what is the magnitude of the average force exerted on the ball by the floor?

600N

Explanation:

If the ball bounces off upward with the same speed, the velocities before and after are equal and in opposite directions. That means the floor must have caused a momentum impact to change its direction.

By the law of momentum conservation:

where m = 3 kg is the ball mass. v = -12 m/s is the velocity right before the impact. V = 12m/s is the velocity right after the impact. ΔP_f is the momentum caused by the floor.

Then the average force exerted during the 0.12s impact is

the answer is a 3.8cm

explanation:

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft