Question

In a ground to ground projection, a projectile has a range of $40\sqrt{3}m$ and reaches a maximum height of 10 m. The angle at which the projectile is fired with horizontal is

• A. $60$
• B. $40$
• C. $30$
• D. $0$

Answer 1

Answer: (B)

$40$

Given,

$R=40\sqrt{3}m$

$H=10m$

$g=10m/s^2$

The horizontal range in projectile motion,

$R=\frac{u^{2}sin2\theta }{g}$

$40\sqrt{3}=\frac{u^{2}sin2\theta }{10}$

$u^{2}sin2\theta =400\sqrt{3}$

$2u^{2}sin\theta cos\theta =400\sqrt{3}$

$u^{2}sin\theta cos\theta =200\sqrt{3}$. . . . . . . . . . . .(1)

The maximum height in projectile motion,

$H=\frac{u^{2}si{n}^2\theta }{2g}$

$10=\frac{u^{2}si{n}^2\theta }{10}$

$u^{2}si{n}^2\theta =100$

$usin\theta =10$

$u=\frac{10}{sin\theta }$. . . . . . . . . . . . .(2)

Substitute equation (2) in equation (1) , we get

$\frac{100}{si{n}^2\theta }sin\theta cos\theta =200\sqrt{3}$

$cot\theta =2\sqrt{3}$

$tan\theta =\frac{1}{2\sqrt{3}}$

$\theta ={40.89}^0$