Topic: Engineering Mechanics
A stone is projected horizontally from a cliff at $10~m/sec$ and lands on the ground below at 20 m from the base of the cliff. Find the height of the cliff. Use $g=10~m/sec^{2}$
- A. 18 m
- B. 20 m
- C. 22 m
- D. 24 m
Answer: B. 20 m
Explanation: Horizontal distance $= velocity \times time \implies 20 = 10 \times t \implies t = 2~sec$. The initial vertical velocity is 0. Vertical distance (height) $h = ut + \frac{1}{2}at^{2} = 0 + \frac{1}{2}(10)(2)^{2} = 20~m$.