Question

A balloon going upward with a velocity of $12{\mathrm{ms}}^{-1}$ is at a height of $65\mathrm{m}$ from the earth’s surface at any instant. Exactly at this instant, a ball drops from it. How much time will the ball take in reaching the surface of the earth? (Take, $\mathrm{g}=10{\mathrm{ms}}^{-2}$)

• A. $5\mathrm{s}$
• B. $6\mathrm{s}$
• C. $10\mathrm{s}$
• D. $\mathrm{None}\mathrm{of}\mathrm{these}$

Answer 1

The correct option is A

$5\mathrm{s}$

Step 1: Given data

Initial velocity of a ball, $\mathrm{u}=12{\mathrm{ms}}^{-1}$

Height of a ball, $\mathrm{S}=-65\mathrm{m}(\because \mathrm{the}\mathrm{direction}\mathrm{of}\mathrm{ballon}\mathrm{and}\mathrm{ball}\mathrm{are}\mathrm{opposite})$

Acceleration of a ball, $\mathrm{a}=\mathrm{g}=-10{\mathrm{ms}}^{-2}$

Step 2: Calculating the time.

By using the equation of motion, we get:

$$\begin{gathered} \mathrm{S}=\mathrm{ut}+\frac{1}{2}{\mathrm{at}}^2 \\ \Rightarrow -65=12\mathrm{t}+\frac{1}{2}\times \left(-10\right)\times {\mathrm{t}}^2 \\ \Rightarrow -65=12\mathrm{t}-5{\mathrm{t}}^2 \\ \Rightarrow 5{\mathrm{t}}^2-12\mathrm{t}-65=0 \\ \Rightarrow 5{\mathrm{t}}^2+13\mathrm{t}-25\mathrm{t}-65=0 \\ \Rightarrow \mathrm{t}\left(5\mathrm{t}+13\right)-5\left(5\mathrm{t}+13\right)=0 \\ \Rightarrow \left(\mathrm{t}-5\right)\left(5\mathrm{t}+13\right)=0 \\ \Rightarrow \mathrm{t}=5,\mathrm{t}=\frac{-13}{5} \end{gathered}$$

But time cannot be negative, hence $\mathrm{t}=5\mathrm{s}$

Therefore option(A) is correct answer.