a body is projected horizonatlly from the top of a cliff 45m above the ground. if the body lands at a distance 30m from the foot of the cliff, calculate the speed of projection. [g = 10ms-2]
A
10ms-1
B
15ms-1
C
20ms-1
D
30ms-1
Verified Correct Answer: A.
Physical Principles & Numerical Solution
from, H = \(\frac{1}{2}\)gt\(^2\)
45 = \(\frac{1}{2}\) x 10 x t\(^2\)
45 = 0.5 x 10 t\(^2\)
45 = 5t\(^2\)
t\(^2\) = \(\frac{45}{5}\) = 9
t = \(\sqrt{9}\) = 3s
But distance = velocity x time
V = \(\frac{D}{t}\) = \(\frac{30}{3}\) = 10m/s
Source: Myschool Past Questions Archive (Record #35273)