Question

A particle is projected vertically upwards from a point $A$ on the ground. It takes $t_1$ time to reach a point $B$ but it still continues to move up. If it takes further $t_2$ time to reach the ground from point $B$ then height of point $B$ from the ground is

• A. $\frac{1}{2}{(t_1+t_2)}^2$
• B. $g{t}_1{t}_2$
• C. $\frac{1}{8}{(t_1+t_2)}^2$
• D. $\frac{1}{2}g{t}_1{t}_2$

Answer 1

The correct option is D $\frac{1}{2}g{t}_1{t}_2$

Time of flight $=t_1+t_2$

$s=u{t}_1-\frac{1}{2}g{t}_1^2$ [$u$-initial speed, $s$-height of point B from ground]

At maximum height,

$v=0,u=at,t=\frac{u}{a}=\frac{u}{g}$

Time of flight $=2\times \frac{u}{g}=\frac{2u}{g}$

$\frac{2u}{g}=t_1+t_2$

$u=\frac{g(t_1+t_2)}{2}$

$s=\frac{g(t_1+t_2)}{2}\times t_1-\frac{1}{2}g{t}_1^2$

$=\frac{1}{2}g{t}_1^2+\frac{1}{2}g{t}_1{t}_2-\frac{1}{2}g{t}_1^2$

$=\frac{1}{2}g{t}_1{t}_2$