a ball is released from a height of 10m . if it loses 20% of its energy on hitting the ground, the height to which it bounces is? (ans : 8m )
Hi,
Please find below the solution to the asked query:
Let m be the mass of the ball and it is fallen from the height h.
From the principal of conservation of energy, we have
Kinetic energy gained by the ball in falling down = Potential energy lost by the ball in falling down = mgh
We have given that the ball loses 20 % on stricking. Therefore, the ball will rise up with 80% energy.
Let h' be the height to which the ball rises.
mgh’ = 80% of mgh
⇒ mgh’ = 0.80 mgh
⇒ h’ = 0.80 h = 0.80 x 10 = 8 m
Hope this information will clear your doubts.
Please find below the solution to the asked query:
Let m be the mass of the ball and it is fallen from the height h.
From the principal of conservation of energy, we have
Kinetic energy gained by the ball in falling down = Potential energy lost by the ball in falling down = mgh
We have given that the ball loses 20 % on stricking. Therefore, the ball will rise up with 80% energy.
Let h' be the height to which the ball rises.
mgh’ = 80% of mgh
⇒ mgh’ = 0.80 mgh
⇒ h’ = 0.80 h = 0.80 x 10 = 8 m
Hope this information will clear your doubts.