Question

Find the value of $\theta$ in the diagram, so that the projectile can hit the target, placed at a height of $10m$ from ground (as shown). $(g=10{\text{m/s}}^2$)

• A. $\theta ={30}^0$
• B. $\theta ={45}^0$
• C. $\theta ={\tan }^{-1}5$
• D. $\theta ={80}^0$

Answer 1

The correct option is B $\theta ={45}^0$

When the projectile (projected with initial speed $(u=20\text{m/s})$ hits the target, then its $x$ & $y$ co-ordinates should be equal to that of target.

Hence $x=20\text{m}$ and $y=10\text{m}$, now we will substitute it in the equation of trajectory, which is given as;
$y=x\tan \theta -\frac{g{x}^2}{2u^2{\cos }^2\theta }$

$10=20\times \tan \theta -\frac{10\times {20}^2}{2\times {20}^2}\times \left({\sec }^2\theta \right)$

$2=4\times \tan \theta -(1+{\tan }^2\theta )$

${\tan }^2\theta -4\tan \theta +3=0$

$\tan \theta =3,1$
Therefore, corresponding to these values of $\tan \theta$,
either $\theta ={45}^{\circ }$ or ${\tan }^{-1}\left(3\right)$

Hence option B is the correct answer