BITSAT 2015 Paper

Let the body is projected vertically upwards from A with a speed $u_0$ Using equation, s = ut + $\frac{1}{2}at^2$ For case (1) – h = $u_0 t_2 - \frac{1}{2} g t_1^2$ ... (1) For case (2) – h = $u_0 t_2$ - $\frac{1}{2} g t_1^2$ ... (2) Subtracting eq (2) from (1), we get 0 = $u_0 (t_2 + t_1)$ + $\frac{1}{2} g (t_2^2 - t_1^2)$ ⇒ $u_0$ = $\frac{1}{2} g (t_1 - t_2)$ ... (3) Putting the value of $u_0$ in eq (2), we get - h = - $\left(\frac{1}{2}\right)$ g $(t - 1 - t_2) t_2$ - $\frac{1}{2} g t_2^2$ ⇒ h = $\frac{1}{2} g (t_1 t_2)$ ... (4) For case 3, $u_0$ = 0, t = ? - h = 0 × t - $\frac{1}{2} g t^2$ h = $\frac{1}{2} g t^2$ Comparing eq. (4) and (5), we get $\frac{1}{2} g t^2$ = $\frac{1}{2} g t_1 t_2$ ∴ t = $\sqrt{t_1 t_2}$