Question

A ball is dropped from the top of a $100\mathrm{m}$ high tower on a planet. In the last $1/2\mathrm{s}$ before hitting the ground, it covers a distance of $19\mathrm{m}$ . Acceleration due to gravity (in ${\mathrm{ms}}^{-2}$) near the surface on that planet is

Answer 1

Step 1: Given data

A ball is dropped from a $100\mathrm{m}$ high tower.

Let the ball covers the distance of ${\mathrm{S}}_1=81\mathrm{m}$ in $\mathrm{t}$ sec.

And the ball covers the distance of ${\mathrm{S}}_2=100\mathrm{m}$ in $\left(\mathrm{t}+\frac{1}{2}\right)$ sec.

Initial velocity of a ball, $\mathrm{u}=0$

Step 2: Finding the acceleration due to gravity.

Here, use the equation of motion, for ${\mathrm{S}}_1=81\mathrm{m}$

$$\begin{gathered} {\mathrm{S}}_1=\mathrm{ut}+\frac{1}{2}{\mathrm{at}}^2 \\ \Rightarrow 81=0+\frac{1}{2}{\mathrm{at}}^2 \\ \Rightarrow \mathrm{t}=9\sqrt{\frac{2}{\mathrm{a}}}\left(1\right) \end{gathered}$$

For ${\mathrm{S}}_2=100\mathrm{m}$,

$$\begin{gathered} {\mathrm{S}}_2=\frac{1}{2}\mathrm{a}{\left(\mathrm{t}+\frac{1}{2}\right)}^2 \\ \Rightarrow 100=\frac{1}{2}\mathrm{a}{\left(\mathrm{t}+\frac{1}{2}\right)}^2 \\ \Rightarrow \mathrm{t}+\frac{1}{2}=10\sqrt{\frac{2}{\mathrm{a}}} \end{gathered}$$

Put the value of $\mathrm{t}$ from equation first we get,

$$\begin{gathered} 9\sqrt{\frac{2}{\mathrm{a}}}+\frac{1}{2}=10\sqrt{\frac{2}{\mathrm{a}}} \\ \Rightarrow \frac{1}{2}=\sqrt{\frac{2}{\mathrm{a}}} \\ \Rightarrow \mathrm{a}=8{\mathrm{ms}}^{-2} \end{gathered}$$

Therefore, the value of acceleration due to gravity near the surface of the planet is $\mathrm{a}=8{\mathrm{ms}}^{-2}$