Question

a stone is thrown vertically upwards with initial velocity u from the top of the tower reaches the ground with a velocity 3 u the height of the Tower is​

Answer 1

Explanation:

Use energy conservation

$\frac{1}{2} m {u}^{2} + mgh = \frac{1}{2} m {(3u)}^{2} \\ gh = ( \frac{9 - 1}{2} ) {u}^{2} \\ h = \frac{4 {u}^{2} }{g}$