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

Answer 1

Answer: (D)

Total time of particle $=t_1+t_2$
to find initial value
Applying $2^{\text{nd }}$ equation of motion
$o=u\left(t_1+t_2\right)-\frac{g}{2}{\left(t_1+t_2\right)}^2$
$\Rightarrow \frac{g}{2}\left(t_1+t_2\right)=u[\because$ height at $t=t_1+t_2$ is zero $]$
At $t=t_1$, height attained is 1
So,
$<br/>\begin{array}{l}<br/>h=u{t}_1-\frac{g}{2}{t}_1^2\\ <br/>h=\frac{g}{2}\left(t_1+t_2\right)t_1-\frac{g}{2}{t}_1^2\\ <br/>h=\frac{g}{2}{t}_1\left[t_1+t_2-t_1\right]\\ <br/>h=\frac{g}{2}{t}_1{t}_2<br/>\end{array}<br/>$