A ball is thrown vertically downward with a velocity $u$ from the top of a tower. It strikes the ground with a velocity $3 u$ . The time taken by the ball to reach the ground is given by

\frac{u}{g}

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\frac{2 u}{g}

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\frac{3 u}{g}

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\frac{4 u}{g}

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Solution

The correct option is B $\frac{2 u}{g}$
Using first equation of motion, we have
$v = u + a t$ ,
As $v = 3 u$ and $a = g$ , we have
$3 u = u + g t \Rightarrow 2 u = g t \Rightarrow t = \frac{2 u}{g}$

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A ball is thrown vertically downward with a velocity u from the top of a tower. It strikes the ground with a velocity 3u. The time taken by the ball to reach the ground is given by

The correct option is B 2ugUsing first equation of motion, we have v=u+at, As v=3u and a=g, we have 3u=u+gt⇒2u=gt⇒t=2ug

A ball is thrown vertically downward with a velocity $u$ from the top of a tower. It strikes the ground with a velocity $3 u$ . The time taken by the ball to reach the ground is given by

The correct option is B $\frac{2 u}{g}$
Using first equation of motion, we have
$v = u + a t$ ,
As $v = 3 u$ and $a = g$ , we have
$3 u = u + g t \Rightarrow 2 u = g t \Rightarrow t = \frac{2 u}{g}$