Question

If a object of mass m is thrown upward from sjrface of earth with initial velocity u and it travelled it course and fell back to ground. While going upward or downward at a certain point it's kinetic and potential energy is equal.If the velocity is doubled then find the ratio of kinetic and potential energy of that same point ?​Please send the solution.​

Answer 1

Answer:

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Answer 2

Answer:

3

Explanation:

In first case 1/2mV^2 = mgh + 1/2mv^2 now at height h 1/2mv^2= mgh

Using this we get 1/2mV^2=2(1/2mv^2)

V=v√2...(1)

ACC to question

1/2m(2V)^2= mgh + 1/2m(v1)^2

mgh=1/2mv^2

2V^2= ( V/√2)^2+ (v1)^2

Ratio = (v1/v)^2= 3