A ball is thrown vertically upward with a velocity $μ$ from the top of a tower. If it strikes the ground with velocity $3 μ$ , the time taken by the ball to reach the ground is:

\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 D $\frac{4 u}{g}$
Given,

Initial velocity, $u = - μ m s^{_{- 1}}$

Final velocity, $v = 3 μ m s^{- 1}$

Apply first kinematic equation of motion

$v = u + a t$

$3 μ = - μ + g t$

$t = \frac{4 μ}{g}$

Hence, time required to reach ground is $\frac{4 μ}{g}$

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A ball is thrown vertically upward with a velocity μ from the top of a tower. If it strikes the ground with velocity 3μ, the time taken by the ball to reach the ground is:

The correct option is D 4ugGiven, Initial velocity, u=−μms−1 Final velocity, v=3μms−1 Apply first kinematic equation of motion v=u+at 3μ=−μ+gt t=4μg Hence, time required to reach ground is 4μg

A ball is thrown vertically upward with a velocity $μ$ from the top of a tower. If it strikes the ground with velocity $3 μ$ , the time taken by the ball to reach the ground is:

The correct option is D $\frac{4 u}{g}$
Given,

Initial velocity, $u = - μ m s^{_{- 1}}$

Final velocity, $v = 3 μ m s^{- 1}$

Apply first kinematic equation of motion

$v = u + a t$

$3 μ = - μ + g t$

$t = \frac{4 μ}{g}$

Hence, time required to reach ground is $\frac{4 μ}{g}$