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A body projected vertically upwards with a velocity $ u $ returns to the starting point in 4 seconds. If $ g = 10m{s^{ - 2}} $ the value of $ u $ is (m/s)
(A) 5
(B) 20
(C) 40
(D) 10

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Last updated date: 21st Jun 2024
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Answer
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Hint At the maximum height, the magnitude of the velocity of the stone is zero. The time taken during ascent is equal to half the time taken for the total journey. So from the equation of motion, we can calculate the initial velocity.
Formula used: In this solution we will be using the following formulae;
 $ s = ut \pm \dfrac{1}{2}g{t^2} $ where $ s $ is the total distance covered during travel. $ u $ is the initial velocity at which the stone was thrown, $ g $ is the acceleration due to gravity and $ t $ is the time taken.
 $ v = u \pm gt $ where $ v $ is the final velocity

Complete step by step answer
A body thrown vertically descends through the same path at which it ascends. And at the negligence of air resistance, the total time taken for ascent is half the total time taken for entire journey, hence $ {t_a} = \dfrac{1}{2}{t_t} $ where $ {t_a} $ is the time taken to ascend to maximum height and $ {t_t} $ is the total time taken to ascend and descend. Thus,
 $ {t_a} = \dfrac{1}{2}\left( 4 \right) = 2s $
Hence, from $ v = u - gt $ where $ v $ is the final velocity, $ u $ is the initial velocity at which the stone was thrown upward and $ g $ is the acceleration due to gravity, at maximum height, we have
 $ 0 = u - g{t_a} $
Making $ u $ subject of the formula, by adding $ - g{t_a} $ to both sides, we have
 $ u = g{t_a} $
Inserting all known values, we have
 $ u = 10\left( 2 \right) = 20m/s $
Hence, the correct option is B.

Note
Alternatively, from the knowledge that air resistance is neglected, the magnitude of velocity of descent at starting point (since end point is the same as starting point) is the magnitude of velocity of ascent at starting point.
Hence, we have from $ v = u - gt $ that
 $ - u = u - g{t_t} $ (upward is taken positive) since $ v = - u $ . $ {t_t} $ is the total time.
Making $ u $ subject of the formula we have
 $ 2u = g{t_t} $ . Hence
 $ u = \dfrac{{g{t_t}}}{2} $
Inserting values we have
 $ u = \dfrac{{10\left( 4 \right)}}{2} = 20m/s $ .