A stone is thrown upwards with a velocity $v$ from the top of a tower. It reaches the ground with a velocity $3 v$ . What is the height of the tower?

\frac{2 v^{2}}{g}

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\frac{3 v^{2}}{g}

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\frac{4 v^{2}}{g}

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\frac{6 v^{2}}{g}

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Solution

The correct option is B $\frac{4 v^{2}}{g}$
$v^{2} - u^{2} = 2 a s \Rightarrow {(3 v)}^{2} - v^{2} = 2 g h \Rightarrow h = \frac{{4 v}^{2}}{g}$

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(g=10m/s)

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