Question

A ball of mass 1 kg dropped from 9.8 m height strikes the ground and rebounds to a height of 4.9m if the time of contact between ball and ground is 0.1s then find impulse and average force acting on ball

Answer 1

》》Answer 《《

Velocity when it hits the ground

u = sqrt(2gH)

= sqrt(2 * 9.8 * 9.8)

= 9.8 sqrt(2) m/s

Velocity just after hitting the ground

v = sqrt(2gH)

= sqrt(2 * 9.8 * 4.9)

= 9.8 m/s

Impulse = change in momentum

= mass × change in velocity

= m * (v - u)

= 1kg * (9.8 - (-9.8*sqrt(2)) m/s

= 9.8 * (1 + sqrt(2)) kgm/s

= 9.8 * 1.414 kgm/s

= 13.8572 kgm/s

Average Force = Impulse / Time taken

= 13.8572 / 0.1

= 1.38 N

Answer 2

Answer:

the answer will be 9.8sqrt