Question

An object is projected upwards from the foot of a tower. The object crosses the top of the tower twice with an interval of 8 s and the object reaches the foot after 16 s. The height of the tower is (Given \( g = 10 \, m/s^2 \))

• A. \( 220 \) m
• B. \( 240 \) m
• C. \( 640 \) m
• D. \( 80 \) m

Hint:

For vertically projected motion, the time of flight is twice the time taken to reach the highest point. Use \( h = \frac{1}{2} g T^2 \) to determine the height.

Answer 1

The Correct Option is B

Step 1: Understanding motion and time intervals 
Given that the object crosses the top of the tower twice with an interval of 8 s and reaches the ground after 16 s, we interpret this as a symmetric motion: - The total time of flight: \( T = 16 \) s. - Time taken to reach maximum height: \( T/2 = 8 \) s. 
Step 2: Using kinematic equations 
We use the equation of motion: \[ h = \frac{1}{2} g T^2 \] Substituting values: \[ h = \frac{1}{2} \times 10 \times \left( \frac{16}{2} \right)^2 \] \[ h = \frac{1}{2} \times 10 \times 64 \] \[ h = 5 \times 64 = 240 { m}. \] 
Step 3: Conclusion 
Thus, the height of the tower is: \[ 240 { m}. \]