Question

A body thrown up with a certain velocity reaches a maximum height h. At a
point p above the ground. P.E: K.E = 1:2. If the same body is thrown with
double the velocity, then the ratio of P.E and K.E at the same point p is
(Ans: 1:11)​

Answer 1

If the same body is thrown with  double the velocity, then the ratio of P.E and K.E at the same point p is  1:11

Explanation:

As we know that body is projected upwards such that it will reach to maximum height of "h"

So the total mechanical energy at the top position is given as

$mgh = KE + PE$

also we know that

PE : KE = 1 : 2

so we have

$PE = \frac{mgh}{3}$

$KE = \frac{2mgh}{3}$

Now we know that the speed is doubled

So the we have

total mechanical energy is given as

$ME = \frac{1}{2}m(2v_i)^2$ = 4mgh[/tex]

now again at same height

$PE = \frac{mgh}{3}$

so by mechanical energy conservation we have

$4mgh = \frac{1}{3}mgh + KE$

$KE = \frac{11}{3} mgh}$

so the ratio of PE and KE is

PE : KE = 1 : 11

#Learn

Topic : Mechanical energy conservation law

https://brainly.in/question/6270071