Question

A cannon fires successively two shells with velocity $v_0=250m/s$; the first at the angle ${\theta }_1={60}^{\circ }$ and the second at the angle ${\theta }_2={45}^{\circ }$ to the horizontal, the azimuth being the same. Neglecting the air drag, the time interval(in seconds) between firings leading to the collision of the shells is (10+x). Find the value of x. (Take $g=10m/s^2$ and round-off your answer to the nearest integer.)

Answer 1

Let the flight time of the shell 1 and shell 2 be $t_1{andt}_2$ . If they both are going to collide then the coordinates of both the shells must be same.

1. Horizontal displacement

Displacement formula - $S_{horizontal}=u_{horizontal}t+\frac{1}{2}{a}_{horizontal}{t}^2$

$⟹ucos\left({60}^{\circ }\right)\times t_1=ucos\left({45}^{\circ }\right)\times t_2$

$⟹t_1=t_2\sqrt{2}...\left(1\right)$

2. Vertical Displacement

Displacement formula - $S=u_{vertical}t+\frac{1}{2}{a}_{vertical}{t}^2$

$⟹usin\left({60}^{\circ }\right)t_1-\frac{1}{2}g{t_1}^2=usin\left({45}^{\circ }\right)t_2-\frac{1}{2}g{t_2}^2$

$⟹u\frac{\sqrt{3}}{2}{t}_1-\frac{1}{2}g{t_1}^2=\frac{u}{\sqrt{2}}{t}_2-\frac{1}{2}g{t_2}^2...\left(2\right)$

Now after solving equation 1 and equation 2 we get the result:

$⟹t_2=\frac{u\sqrt{2}(\sqrt{3}-1)}{g}$

$⟹t_1=\frac{2u(\sqrt{3}-1)}{g}$

Therefore, the time interval is:
$⟹(t_1-t_2)=\frac{u(\sqrt{3}-1)}{g}(2-\sqrt{2})$

$⟹10.72sec$

From the equation given in the question:

$⟹(10+x)=10.72$

Nearest Integer is 1