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)

$\frac{1}{2} gt_1t_2$

Answer 2

Time taken for the particle to reach the highest point is $\frac{t _{1}+ t _{2}}{2}.$ As $\upsilon = u - gt$ At highest point, $\upsilon = 0$ Therefore, initial velocity of the particle is $u = g\left(\frac{t _{1}+ t _{2}}{2}\right)$ $\quad\ldots\left(i\right)$ Therefore, height of point B from the ground is $h = ut_{1} - \frac{1}{2}gt^{2}_{1} = g \left(\frac{t _{1}+ t _{2}}{2}\right)t_{1} - \frac{1}{2}gt^{2}_{1}\quad$ (Using (i)) or $h = g \left(\frac{t^{2}_{1}}{2}+\frac{t_{1}t_{2}}{2}\right) - \frac{1}{2}gt^{2}_{1}$ or $h = \frac{1}{2}g t_{1}t_{2}$