Question

# A stone is dropped from a tower at a height of $h$. Another stone is thrown up simultaneously from the ground, which can reach a maximum height of $4h$. Determine the time at which both the stones meet.

Open in App
Solution

## Step 1: Given data${h}_{1}=h$${h}_{2}=4h$${u}_{1}=0$Step 2: To find Time at which both the stones meet.Step 3: Calculate the time of the meeting:Use the second equation of motion to find the height, $h$.$s=ut+\frac{1}{2}a{t}^{2}$Let the initial velocity of the stone that is thrown from the ground ${u}_{2}$​. At maximum height $4\text{h}$, Velocity ${v}_{2}=0$${{v}_{2}}^{2}={{u}_{2}}^{2}-2gs\phantom{\rule{0ex}{0ex}}{0}^{2}={{u}_{2}}^{2}-2g\left(4h\right)\phantom{\rule{0ex}{0ex}}{u}_{2}=\sqrt{8gh}$Let ${h}_{1}$be the distance covered by the stone which is dropped from height $h$ and ${h}_{2}$ be the distance covered by the stone which is thrown up from the ground. Let time $t$ both stones crossed each other.${h}_{1}=\frac{1}{2}g{t}^{2}...\left(1\right)\left[\because {u}_{1}=0\right]$ ${h}_{2}=\left(\sqrt{8gh}\right)t-\frac{1}{2}g{t}^{2}\phantom{\rule{0ex}{0ex}}{h}_{2}=\sqrt{8gh}t-\frac{1}{2}g{t}^{2}...\left(2\right)$ Adding ${h}_{1}$​ and ${h}_{2}$ gives the height $h$.Add equations (1) and (2).${h}_{1}+{h}_{2}=\sqrt{8gh}t\phantom{\rule{0ex}{0ex}}h=\sqrt{8gh}t\phantom{\rule{0ex}{0ex}}t=\sqrt{\frac{h}{8g}}$Hence, the time at which both the stones meet is $\sqrt{\frac{h}{8g}}$.

Suggest Corrections
0