Question

A stone is dropped from a balloon going up with a uniform velocity of $5.0\text{m/s}$. If the balloon was $50\text{m}$ high when the stone was dropped, find its height when stone hits the ground.

• A. $50\text{m}$
• B. $68.5\text{m}$
• C. $78.5\text{m}$
• D. $40\text{m}$

Answer 1

The correct option is B $68.5\text{m}$

At $\text{t}=\text{0}$, the stone was going up with a velocity of $5\text{m/s}$. After that it moved as a freely falling particle with downward acceleration $\text{g}$. Take vertically upward as the positive $\text{X-}$axis. If it reaches the ground at time $\text{t}$,

$\text{x}$$=\text{-50 m}$,
$\text{u}$$=\text{5 m/s}$,
$\text{a}$$={\text{-10 m/s}}^2$,

We have $\text{x}$ $=\text{ut+}$$\frac{1}{2}\text{a}{\text{t}}^2$
or, $\text{-50 m}=\text{(5 m/s)}\text{t}+\frac{1}{2}{\text{(-10 m/s}}^2)\times {\text{t}}^2$
or, $\text{t}$$=\frac{1\pm \sqrt{41}}{2}\text{s}$
or, $\text{t}$$=\text{-2.7 s}$ or $,\text{t}$$=\text{3.7 s}$
Negative $\text{t}$has no significance in the problem. The stone reaches the ground at $\text{t}=\text{3.7 s}$. During this time the balloon has moved uniformly up. The distance covered by it is
$=\text{(5 m/s)}$$\text{.(3.7 s)}$$=\text{18.5 m}$
Hence, the height of the balloon when the stone reaches the ground is $\text{50 m}+\text{18.5 m}$$=\text{68.5 m}$