Question

A body projected vertically upwards crosses a point at a height $h$ twice in its journey at instants $t_{1}$ and $t_{2}$ second. The maximum height reached by the body is

• A. $\frac{g}{4}\left(t_{1} + t_{2}\right)^{2}$
• B. $g\left(\frac{t_{1} + t_{2}}{4}\right)^{2}$
• C. $2g\left(\frac{t_{1} + t_{2}}{4}\right)^{2}$
• D. $\frac{g}{4}\left(t_{1} t_{2}\right)$

Answer 1

Answer: (C)

Time taken by the body to reach the point $A$ is $t_{1}$ (During upward journey).

The body crosses this point again (during the downward journey) after $t_{2}, \, ie, \, $ the body takes the time $\left(t_{2} - t_{1}\right)$ to come again at the point $A.$
So, the time taken by the body to reach at point $B$ (a maximum height).
$t=t_{1}+\left(\frac{t_{2} - t_{1}}{2}\right)$
$\left[\because T i m e \, o f \, a s c e n d i n g = T i m e \, o f \, d e s c e n d i n g\right]$
$t=\frac{t_{1} + t_{2}}{2}$
So, maximum height $H=\frac{1}{2} \, gt^{2}$
$ \, =\frac{1}{2}g\left(\frac{t_{1} + t_{2}}{2}\right)^{2}$
$ \, \, \, =2g\left(\frac{t_{1} + t_{2}}{4}\right)^{2}$