Question

A stone is thrown vertically at a speed of $30m$ $s^{-1}$ making an angle of $45^{\circ }$ with the horizontal. What is the maximum height reached by the stone ? Take $g=10m{s}^{-2}$.

• A. 22.5 m
• B. 10 m
• C. 30 m
• D. 15 m

Answer 1

Answer: (A)

Given

$v=30m{s}^{-1},\theta =45^{\circ }$
$g=10m{s}^{-2},H=?$ (maximum height)
Maximum height of the projectile moving with velocity $v$ at an angle $\theta$ is given by
$H=\frac{v^2{\sin }^2\theta }{2g}$
$=\frac{30^2\times {\sin }^2(45)}{2\times 10}$
$=900\times {\left(\frac{1}{\sqrt{2}}\right)}^2/20=\frac{450}{20}$
$=22.5m$