Question

A particle is projected at an angle ${30}^{\circ }$ with the horizontal at speed $20m/s$. How high will it strike a wall $8\sqrt{3}m$ away from point of projection?
$(g=10m/s^2)$

• A. $5m$
• B. $4.8m$
• C. $2.4m$
• D. $9.6m$

Answer 1

The correct option is B $4.8m$

$x=8\sqrt{3},\theta ={30}^{\circ }R=\frac{u^2\sin 2\theta }{g}$
$=\frac{u^2\times 2\times \sin \theta \times \cos \theta }{g}$
$=\frac{{20}^2\times 2\times \frac{1}{2}\times \frac{\sqrt{3}}{2}}{10}=20\sqrt{3}$
By equation of trajectory of a projectile, we know
$y=x\tan \theta (1-\frac{x}{R})\Rightarrow y=8\sqrt{3}\times \frac{1}{\sqrt{3}}(1-\frac{8\sqrt{3}}{20\sqrt{3}})=8\times (1-\frac{2}{5})=8\times \frac{3}{5}=4.8m$