Question

A projectile is aimed at a mark on a horizontal plane through the point of projection and falls $6$m short when its elevation angle is ${30}^o$. The projectile overshoots the mark by $9$m when the angle of projection is ${45}^o$. The correct distance of the mark is?

• A. $\frac{(9\sqrt{3}+12)}{2-\sqrt{3}}$m
• B. $(30\sqrt{3}+51)m$
• C. $(60+30\sqrt{3})m$
• D. $30(2-\sqrt{3})m$

Answer 1

The correct option is A $\frac{(9\sqrt{3}+12)}{2-\sqrt{3}}$m

We know horizontal range of a projectile $R=\frac{v_0^2\sin 2\theta }{g}$

Now range of projectile when $\theta ={30}^0$ gets short by 6 m from the target,thus range of projectile

$R-6=\frac{v_0^2\sin 60}{g}$-------(1) and

range of projectile when $\theta ={45}^0$ gets overshoot by 9 m from the target,thus range of projectile

$R+9=\frac{v_0^2\sin 90}{g}$--------(2)

Considering the speed of the projectile to be same for both the condition we can equate 1 and 2 by

$\frac{(R-6)g}{sin{60}^0}$=$\frac{(R+9)g}{sin{90}^0}2R-12=R\sqrt{3}+9\sqrt{3}R(2-\sqrt{3})=9\sqrt{3}+12R=\frac{9\sqrt{3}+12}{2-\sqrt{3}}$

$R=19.85m$