Question

pls help me .......​

Answer 1

I) kinetic energy = 1/2 x mass x (velocity)^2 = 1/2 x 1kg x (20m/s)^2 = 200 J

ii) Height from the ground be s
From position velocity relation,
V^2 - u^2 = 2gs
(20m/s)^2 - 0 ^2 = 2 x 10m/s^2 x s Initial velocity is zero because the ball is dropped
400 = 20s
S = 20m
Therefore height from the ground is 20m (I)
Now,
Potential energy = mass x g x height = 1kg x 10m/s^2 x 20m [ from (I)]
= 200 J

iii) height is 20m [from(I)]