Question

the ball of mass 1000g is dropped from a height of. 5m and rebounds to a height of 25m calaulate ; its kinetic just before impact .2 its initial velocity and kinetic energy​

Answer 1

Step-by-step explanation:

v²=(u²+2gh) ...(i)

So after falling from height 5m=>

u=0

g=10m/s-2

h=5m

Putting all these in equation (i)

v=10m/s

Therefore=>

Kinetic energy just before impact= mv²/2= 1*100/2J

Kinetic energy just before impact=50J

Initial velocity=0m/s

Initial kinetic energy=0J