Question

A ball is dropped and it hits the floor at a point A
.It rebounds upwards to a point B while moving from A to B its

Answer 1

Answer:

AT point A the body has kinetic energy only while at B it has potential energy only.

From work-energy theorem:

K

B

−K

A

=V

A

−V

B

0−K

A

=0−V

B

Kinetic energy at A=potential energy at B

Answer 2

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Height from which the ball is dropped is, h = 10 m

Velocity with which the ball hits the ground can be found as,

v^2 = u^2 + 2gh

=> v^2 = 2gh

=> v = (2gh)^1/2 [downward]

The ball then rebounds to a height of, h/ = 2.5 m. Let the velocity with which the ball rebounds be v/.

So, using,

02 = (v/)^2 – 2gh/

=> v/ = (2gh/)^1/2 [upward]

The time for which the ball was in touch with the ground is t = 0.01 s

So, acceleration of the ball is,

a = (v/ - v)/t

[considering upward velocity to be positive, v/ is positive and v is negative]

=> a = [(2gh/)^1/2 – {-(2gh)^1/2}]/t

=> a = [(2×9.8×2.5)^1/2 + (2×9.8×10)^1/2]/0.01

=> a = 2100 m/s2