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Physics Question on Motion in a straight line

A man throws a ball vertically upward and it rises through 20 m and returns to his hands. What was the initial velocity (u) of the ball and how much time (T) is remained in the air? (Take g=10 m s−2g = 10\, m\, s^{-2})

A

u=10 m s−1, T=2 su = 10\, m\, s^{-1},\, T = 2\, s

B

u=10 m s−1, T=4 su = 10\, m\, s^{-1},\, T = 4\, s

C

u=20 m s−1, T=2 su = 20\, m\, s^{-1},\, T = 2\, s

D

u=20 m s−1, T=4 su = 20\, m\, s^{-1},\, T = 4\, s

Answer

u=20 m s−1, T=4 su = 20\, m\, s^{-1},\, T = 4\, s

Explanation

Solution

Taking vertically upward motion of ball, we have υ2=u2+2gs\upsilon^{2} = u^{2} + 2gs or 0=u2+2×(−10)×200 = u^{2} + 2 \times \left(-10\right)\times 20 or u=20 m s−1u = 20\, m \,s^{-1} Taking vertical motion of ball for whole flight, S=ut+12gt2S = ut + \frac{1}{2}gt^{2} 0=20×t+12×(−10)×t20 = 20\times t+\frac{1}{2}\times\left(-10\right)\times t^{2} 5t2−20 t=0?t=4 s.5t^{2} - 20\, t = 0 ? t = 4 \,s.