A stone is projected horizontally from the top of a tower with a speed of 5ms-1. It lands on the ground level at a horizontal distance of 20m from the foot ...

Assessment: WAEC SSCE - Physics - 1999 (Essay) Subject: Physics

Question 1 Report

A stone is projected horizontally from the top of a tower with a speed of 5ms-1. It lands on the ground level at a horizontal distance of 20m from the foot of the tower. Calculate the height of the tower. (g = 10ms\(^{-2}\))

Answer Details

Horizontal projectile. The horizontal and vertical motions are independent. The horizontal velocity stays constant while the vertical motion is free fall from rest.

Step 1: Find the time of flight from the horizontal motion. There is no horizontal acceleration, so

\[ x = ut \implies 20 = 5 \times t \implies t = 4\,\text{s} \]

Step 2: Use the time in the vertical motion. Vertically the stone starts with zero vertical velocity and falls a height \(h\):

\[ h = ut + \tfrac{1}{2}gt^2 = 0 + \tfrac{1}{2}(10)(4)^2 = \tfrac{1}{2}(10)(16) \] \[ h = 80\,\text{m} \]

Answer: the height of the tower is \(80\,\text{m}\).

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