Question 3

A ball is released from the top of a tower of height $$h$$ metres. It takes $$T$$ seconds to reach the ground. What is the position of the ball in $$T/3$$ seconds?

Solution

Solution & Explanation

1. Analyze the Total Journey of the Ball

The ball is dropped from rest at the top of a tower of height $$h$$, meaning its initial velocity ($$u$$) is zero:

$$u = 0 \,\, \text{m/s}$$

Let $$T$$ be the total time taken by the ball to reach the ground. Using the second equation of motion ($$s = u \cdot t + \frac{1}{2} \cdot a \cdot t^2$$) with a downward acceleration due to gravity ($$a = g$$):

$$h = 0 \cdot T + \frac{1}{2} \cdot g \cdot T^2$$

$$h = \frac{1}{2} \cdot g \cdot T^2 \quad \text{--- (Eq. 1)}$$

2. Calculate the Distance Covered in Time $$\frac{T}{3}$$

Let $$y$$ be the downward vertical distance covered by the ball from the top of the tower during the initial time interval $$t = \frac{T}{3}$$:

$$y = 0 \cdot \left(\frac{T}{3}\right) + \frac{1}{2} \cdot g \cdot \left(\frac{T}{3}\right)^2$$

$$y = \frac{1}{2} \cdot g \cdot \frac{T^2}{9}$$

$$y = \frac{1}{9} \cdot \left(\frac{1}{2} \cdot g \cdot T^2\right)$$

Substituting the value of $$h$$ from Equation 1 into this expression gives:

$$y = \frac{h}{9}$$

This represents the distance the ball has fallen downward from the top of the tower.

3. Determine the Position Relative to the Ground

The position of the ball measured upward from the ground ($$h'$$) is equal to the total height of the tower minus the vertical distance it has traveled downward:

$$h' = h - y$$

$$h' = h - \frac{h}{9}$$

$$h' = \frac{9h - h}{9} = \frac{7h}{9} \,\, \text{metres}$$

Concept Check: Because displacement under constant acceleration scales quadratically with time ($$s \propto t^2$$), traveling for one-third of the total time ($$\frac{1}{3} \cdot T$$) means the object covers only one-ninth ($$\frac{1}{9}$$) of the total vertical height from the release point, leaving it at a height of $$\frac{8h}{9}$$ from the ground base.