Question

A projectile is fired vertically up from the surface of earth with a velocity kve, where ve is the escape velocity and k < 1. the maximum height upto which the projectile will rise from the surface of earth is:

Answer 1

Hope it helps

V = k× Ve

Let's write it as V = Ve÷(1÷k)

h (max.height ) = R ÷ (n^2 - 1) . Here n = 1÷k

Hence h = R÷ ((1÷k^2) - 1 )

On simplifing we get , h = (R × k^2 )÷(1- k^2 )

Given k is less than 1. So k^2 can be neglected.

Hence the answer is R ÷ (1 - k^2 )