Question

A stone is projected vertically up with a velocity 'u' reaches a maximum height 'h'. When it is at a height of 3 from the
ground, the ratio of K.E and P.E at that point is (Consider P.E = 0 at the point of projection)
(A) 1:1
(B) 1:2
(C) 1:3
(D) 3:1​

Answer 1

Answer:

c- 1:3

Explanation:

I guess you mistyped the question. I had already solved it and the height given was 3h/4. Considering the height to be 3h/4 first, we need to find Potential and Kinetic Energy so that we can find the ratio.

We have,

P.E= mgh

So P.E when height is 3h/4,

P.E= mg×3h/4 =

$\frac{3mgh}{4}$

Now K.E=

$\frac{1}{2} m {v}^{2}$

from third equation of motion,

${v}^{2} = {u}^{2} + 2as$

when height is h/4

${ v}^{2} = {0}^{2} + 2a \times \frac{h}{4}$

${v}^{2} = \frac{gh}{2}$

Putting v^2 in K.E

We have,

K.E=

$\frac{1}{2} \times m \times ( \frac{gh}{2} )$

=

$\frac{mgh}{4}$

Now,

$\frac{k e}{pe} = \frac{ \frac{mgh}{4} }{ \frac{3mgh}{4} }$

Hence, KE:PE= 1:3.