Question

A projectile has a range of $50m$ and reaches a maximum height of $10m$ calculate the angle at which the projectile is fired.

• A. $ta{n}^{-1}\frac{3}{5}$
• B. $ta{n}^{-1}\frac{2}{5}$
• C. $ta{n}^{-1}\frac{3}{7}$
• D. $ta{n}^{-1}\frac{4}{5}$

Answer 1

The correct option is D $ta{n}^{-1}\frac{4}{5}$

$\text{Step1: Range and maximum height of projectile}$.$\text{[Refer Fig.]}$

Let $u$ be the initial velocity of projection and $\theta$ be the angle of the projectile.

Range of the projectile: $R=\frac{u^2\sin 2\theta }{g}$ $....\left(1\right)$

Maximum height of the projectile: $H=\frac{u^2{\sin }^2\theta }{2g}$ $....\left(2\right)$

$\text{Step2: Angle of projectile calculation}$.

Dividing equation $\left(2\right)$ by $\left(1\right)$

$\Rightarrow$ $\frac{H}{R}=\frac{si{n}^2\theta }{2sin2\theta }$ $=\frac{si{n}^2\theta }{4sin\theta cos\theta }$ $=\frac{\tan \theta }{4}$

$\Rightarrow$ $\frac{10}{50}=\frac{\tan \theta }{4}$

$\Rightarrow$ $\theta ={\tan }^{-1}\left(\frac{4}{5}\right)$

Hence, D is the correct answer.