Question

A stone is dropped from a tower of height $100\text{m}$. At the same time, another stone is thrown upwards at a speed of $25\text{m}{\text{s}}^{-1}$. When and where the two stones would meet? Take $\text{g}=10\text{m}{\text{s}}^{-2}$

• A. Time $=2.5\text{s}$ ; Height $=40\text{m}$
• B. Time $=4\text{s}$ ; Height $=80\text{m}$
• C. Time $=6\text{s}$ ; Height $=50\text{m}$
• D. Time $=7.5\text{s}$ ; Height $=100\text{m}$

Answer 1

The correct option is B Time $=4\text{s}$ ; Height $=80\text{m}$

Given:
The height of the tower, $\text{h}=100\text{m}$
Initial velocity of the second stone, ${\text{u}}_2=25\text{m}{\text{s}}^{-1}$
$\text{g}=10\text{m}{\text{s}}^{-2}$
Initial velocity of the first stone, ${\text{u}}_1=0\text{m}{\text{s}}^{-1}$
Let $\text{t}$ be the time instant at which the two stones would meet/hit each other.
For the falling stone, the distance covered:
${\text{s}}_1={\text{u}}_1\text{t}+\frac{1}{2}{\text{gt}}^2$
$⟹{\text{s}}_1=0+\frac{1}{2}\times 10{\text{t}}^2=5{\text{t}}^2$ ------(1)
For the rising stone, the distance covered:
${\text{s}}_2={\text{u}}_2\text{t}-\frac{1}{2}{\text{gt}}^2$
$⟹{\text{s}}_2=25\text{t}-\frac{1}{2}\times 10{\text{t}}^2=25\text{t}-5{\text{t}}^2$
But, ${\text{s}}_1+{\text{s}}_2=100$
$⟹5{\text{t}}^2+25\text{t}-5{\text{t}}^2=100$
$⟹25\text{t}=100⟹\text{t}=4\text{s}$ ----- (2)
Substitute (2) in (1):
${\text{s}}_1=5\times (4{)}^2=5\times 16=80\text{m}$ --- (3)
$\therefore$ The two stones would meet each other at $\text{t}=4\text{s}$, and at a height of $80\text{m}$.