Question

A steel ball of mass 5kg is dropped from the top of a tower. If it reaches the ground in

2 seconds, find (a) the velocity with which it strike the ground.

(b) the height of the building. (c) the kinetic energy of the ball just before touching

the ground and (d) calculate the force acted on the ground, if it is stopped by the

ground in 0.01 second.​

Answer 1

Explanation:

mass, m = 1kg

height, s = 20m

initial velocity of ball, u = 0m/s

acceleration, a = 10m/s

2

Using, v

2

= u

2

+ 2as

= 0 + 2×10 ×20

= 400

v = 20m/s

Now, let us take upward direction as positive and downward direction as negative.

Initial momentum of the ball (before striking the ground) = m×v

= 1×(-20) (velocity is downward, hence negative)

= -20kgm/s

Now, when the ball strikes and rebounds, its velocity becomes +20m/s (after striking, it moves upward, hence positive velocity)

Final momentum (after striking the ground) = m×v

= 1×(+20)

= +20kgm/s

Change in momentum = Final momentum - Initial momentum

= 20kgm/s - (-20kgm/s)

= 40kgm/s in upward direction.