A ball drops from a height of 25 m. What is the speed attained as it touches the ground?
[g = 10 ms\(^{−2}\)
A
7ms\(^{-1}\)
B
\(\sqrt{3}\)ms\(^{-1}\)
C
10\(\sqrt{5}\)ms\(^{-1}\)
D
20\(\sqrt{5}\)ms\(^{-1}\)
Verified Correct Answer: C.
Physical Principles & Numerical Solution
Height = 25 m, g = 10 ms\(^{−2}\), U = 0m/s
V\(^2\) = U\(^2\) + 2gs
V\(^2\) = 0 + 2 x 10 x 25 = 500
V = \(\sqrt{ 5 \times 100}\) = 10\(\sqrt{5}\)ms\(^{-1}\)
Source: Myschool Past Questions Archive (Record #78272)