A stone thrown vertically upwards attains a maximum height of 45 m. In what time the velocity of stone become equal to one-half the velocity of throw?(Given g = $10 m{s}^{-2}$)

Let us solve the problem in terms of relative initial velocity, relative acceleration and relative displacement of the coin with respect to floor of the lift. Given, $u=0 m{s}^{-1}, a=9.8 m{s}^{-2}, s=4.9 m, t=?$

As $4.9=0\times t+\frac{1}{2}\times 9.8\times t^2$

or $4.9 t^2=4.9$

Or t =1s

15 = 30 -10 t

or 10 t =15

or t =1.5 s