Question

When a body is projected vertically up from the ground it's P.E and K.E at a point p are in the ratio 2:3. if the same body is projected with double the previous velocity at the same point p ratio of its P.E and K.E would be:

Answer 1

Solution

Let ground level be D atam (U=0)

Then, At point P

$U=\Delta U=mgh$

$W_{gravity}=-\Delta U=-mgh$

By work - Energy theorem.

$W_{gravity}=\Delta KE$

$\Rightarrow -mgh=\frac{m{v}^2}{2}-m{u}^2$

$\Rightarrow \frac{m{v}^2}{2}=-mgh+\frac{m{u}^2}{2}$

It is given that $\frac{U}{KE}=\frac{mgh}{\frac{m{u}^2}{2}-mgh}=\frac{2}{3}$

So, $\frac{m{u}^2}{2}=\frac{\overrightarrow{b}mgh}{2}$

If u is doubled, $\frac{m(u^{\prime }{)}^2}{2}=\frac{um{u}^2}{2}=10mgh$

Then at point P,

U = mgh and $\frac{m{v}^{\prime 2}}{2}-\frac{m{u}^{\prime 2}}{2}=-mgh$

$\Rightarrow \frac{m{v}^{\prime 2}}{2}=amgh$

so, $\frac{U}{KE}=\frac{mgh}{\frac{m{u}^{\prime 2}}{2}}=\frac{mgh}{amgh}=1:9$