WAEC - Physics (1994 - No. 33)
A ball is dropped from a height of 45m above the ground. Calculate the velocity of the ball just before it strikes the ground. (Neglect air resistance and take g as \(10ms^{-2}\)
21.2\(ms^{-1}\)
30\(ms^{-1}\)
300\(ms^{-1}\)
450\(ms^{-1}\)
900\(ms^{-1}\)
Explanation
\(v^2 = u^2+2as=0^2 + 2\times 10\times 45 \Rightarrow
v^2 = 900m^2s^{-2} \Rightarrow \\
v = \sqrt{900m^2s^{-2}} \Rightarrow
v = 30ms^{-1}\)
v^2 = 900m^2s^{-2} \Rightarrow \\
v = \sqrt{900m^2s^{-2}} \Rightarrow
v = 30ms^{-1}\)
Comments (0)
