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Q.
A stone is thrown vertically at a speed of $30\, m$ $s^{-1}$ making an angle of $45^\circ$ with the horizontal. What is the maximum height reached by the stone ? Take $g \,= \,10\, ms^{-2}$.
Given
$v=30 ms ^{-1}, \theta=45^{\circ}$
$g=10\, ms ^{-2}, H=?$ (maximum height)
Maximum height of the projectile moving with velocity $v$ at an angle $\theta$ is given by
$H =\frac{v^{2} \sin ^{2} \theta}{2 g}$
$=\frac{30^{2} \times \sin ^{2}(45)}{2 \times 10}$
$=900 \times\left(\frac{1}{\sqrt{2}}\right)^{2} / 20=\frac{450}{20}$
$=22.5 \,m$
