Question

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

• A. $\frac{1}{2}g\left(t _{1}+ t _{2}\right)^{2}$
• B. $gt_1t_2$
• C. $\frac{1}{8}g\left(t _{1}+ t _{2}\right)^{2}$
• D. $\frac{1}{2} gt_1t_2$

Answer 1

Answer: (D)

Total time of particle $=t_{1}+t_{2}$
to find initial value
Applying $2^{\text {nd }}$ equation of motion
$o=u\left(t_{1}+t_{2}\right)-\frac{g}{2}\left(t_{1}+t_{2}\right)^{2}$
$\Rightarrow \frac{g}{2}\left(t_{1}+t_{2}\right)=u\left[\because\right.$ height at $t=t_{1}+t_{2}$ is zero $]$
At $t=t_{1}$, height attained is 1
So,
$
\begin{array}{l}
h=u t_{1}-\frac{g}{2} t_{1}^{2} \\
h=\frac{g}{2}\left(t_{1}+t_{2}\right) t_{1}-\frac{g}{2} t_{1}^{2} \\
h=\frac{g}{2} t_{1}\left[t_{1}+t_{2}-t_{1}\right] \\
h=\frac{g}{2} t_{1} t_{2}
\end{array}
$