Question

A particle is thrown vertically upwards from ground. It takes time $t_1$ to reach a height $h$. It continues to move and takes time $t_2$ to reach the ground. Its maximum height is

• A. $\frac{g}{2}\frac{t_1+t_2}{2}$
• B. $\frac{g}{2}\sqrt{{t_1}^2+{t_2}^2}$
• C. $\frac{g}{8}{(t_1+t_2)}^2$
• D. $g({t_1}^2+{t_2}^2)$

Answer 1

Answer: (C)

$\frac{g}{8}{(t_1+t_2)}^2$

Let $u$ & $H$ be initial velocity and maximum height respectively so, $u=\sqrt{2gH}$
Now, for both time $t_1\&t_2$ displacement is $h$

$h=ut-\frac{1}{2}g{t}^2$

$t^2+\frac{2u}{g}t+\frac{2h}{g}=0$

which is having two roots $t_1$and $t_2$

$t_1+t_2=\frac{2u}{g}or{t}_1+t_2=\frac{2\sqrt{2gH}}{g}\Rightarrow H=\frac{g{(t_1+t_2)}^2}{8}$