Question

A ball is released from the top of a tower of height h meters. It takes T seconds to reach the ground. What is the position of the ball in $\frac{T}{3}sec$

• A. $\frac{h}{9m}fromground$
• B. $\frac{7h}{9m}$from the ground
• C. $\frac{8h}{9m}$ from the ground
• D. $\frac{17h}{18m}$ from the ground

Answer 1

The correct option is C

$\frac{8h}{9m}$ from the ground

Step 1: Given data

Initial velocity u = 0 ms^-1

Acceleration due to gravity = g

In T sec the body travels h meters.

Step 2: Calculations

From the kinematic equation

$$\begin{gathered} S=ut+\frac{1}{2}g{t}^2 \\ h=0+\frac{1}{2}g{t}^2 \\ h=\frac{1}{2}g{t}^2 \end{gathered}$$

at $\frac{T}{3}seconds$ The distance traveled be h₁

$$\begin{gathered} S=ut+\frac{1}{2}g{t}^2 \\ {h}_1=0+\frac{1}{2}g{\left(\frac{T}{3}\right)}^2 \\ {h}_1=\frac{1}{2}g\frac{T^2}{9} \end{gathered}$$

$$\begin{gathered} h_1=\frac{1}{2}g{t}^2\left(\frac{1}{9}\right) \\ {h}_1=\frac{h}{9} \end{gathered}$$

This is the distance from point of release.

Therefore the distance from the ground

.$h-\frac{h}{9}=\frac{8h}{9}$

Hence,

Option ( c) is correct.