Question

Shots are fired from the top of a tower and from bottom simultaneously at angles ${45}^{\circ }$ and ${60}^{\circ }$ respectively as shown. If horizontal distance of the point of collision is at a distance of $(\sqrt{3}+1)$ from the tower,then find the height h of tower (in cm)

Answer 1

We know Range of projectile $R=\frac{v^2\sin 2\theta }{g}$

Given $R=\sqrt{3}+1$ and at $\theta ={60}_0$

$v_1=\sqrt{\frac{Rg}{sin120}}=\sqrt{\frac{\sqrt{3}+1\times 10}{0.867}}=5.61$m/a

And time of flight$T=\frac{2vsin\theta }{g}=\frac{2\left(5.61\right)0.867}{10}=0.975s$

As both the shots falls on the ground at a same range thus its time of flight will remain same and the horizontal

component of the shot will remain same,hence no acceleration on in horizontal direction.

Thus now velocity of second shot $v_2=v_2\cos {45}^0=\frac{v_2}{\sqrt{2}}$

Again $R=\frac{{\left(\frac{v_2}{\sqrt{2}}\right)}^2\sin 2\theta }{g}=\frac{v^2}{2g}$ as $\sin 90=1$

$v=\sqrt{20\left(2.73\right)}=7.38$ m/s

Now taking initial velocity$u=7.38$ m/s time$t=0.975$s and $g=-10m/s^2$

From equation of motion $s=ut+\frac{1}{2}g{t}^2$

$H=7.38\times 0.975-\frac{1}{2}\times 10\times {0.975}^2=2.47$m