Question

when an object is thrown upward,it rises to a height 'h'. how high is the object ,in terms of h, when it has lost one third of it's original kinetic energy?​

Answer 1

New height will be $h'=\frac{2h}{3}$

Explanation:

Let initial potential energy is mgh and initial kinetic energy is $\frac{1}{2}mv^2$

Now according to it has lost $\frac{1}{3}$ of kinetic energy

So left kinetic energy $=KE-\frac{KE}{3}=\frac{2KE}{3}$

According to energy conservation

Kinetic energy will be equal to potential energy

So potential energy $=\frac{2mgh'}{3}$

On comparing with original potential energy new height will be $h'=\frac{2h}{3}$

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Answer 2

Answer:

2h/3

Explanation:

let initial potential enery be mgh and initial K.E be 1/2 mv^2.

if there is lost in K.E is 1/3 then there is still 2/3 of K.E.

left kinetic energy = K.E - K.E/3 = 2K.E/3

SO, a/c to energy conservation.

kinetic energy must be equal to P.E

SO, potential energy = 2m×g×h/3

comparing both equations we get, 2h/3.