Abasketball player drops a 0.60-kg basketball vertically so that it is traveling at 6.3 m/s when it reaches the floor. the ball rebounds upward at a speed of 5.3 m/s. determine the average force that the floor exerts on the ball if the collision lasts 0.12 s.

Explanation:

Force is given by rate of change of momentum.

Mass of ball = 0.60 kg

Initial velocity = -6.3 m/s downwards

Final velocity = 5.3 m/s upwards

Change in momentum = 0.6 x 5.3 - 0.6 x (-6.3)

Change in momentum = 6.96 kg m/s upwards

Time taken = 0.12 s

Rate of change of momentum = Change in momentum ÷ Time

Rate of change of momentum = 6.96 ÷ 0.12

Rate of change of momentum = 58 N upwards

Force = 58 N upwards

The average force acting on the ball is 58 N upwards.

i really dont know but i will try okay.

so i think u would just add all them up and get your answer..

find the sum of the two horizontal vectors and the sum of the two vertical vectors

isoceles right angled triangle