## 1996 (5 marks)

Two guns situated on the top of a hill of height 10 m fire one shot each with the same speed 5$$\sqrt{3}$$ m/s at some interval of time. One gun fires horizontally and other fires upwards at an angle of 60° with the horizontal. The shots collide in air at point P (g = 10 m/s2). Find

(a) the time interval between the firings and
(b) the coordinates of the point P. Take origin of the coordinate system at the foot of the hill right below the muzzle and trajectories in x-y plane

Solution :

Let the projectiles collide at point P, the projectile that was fired next say at time t0 reached at the point of collision after time T i.e. the first projectile took a total time of t0 + T. At time t0 the first projectile was at somewhere between Q and P on its trajectory, let's call this point M (x0,y0). Now we prepare the solution,

The x component of initial velocity of projectile 1, $$u_{x} = u cos 60^{0} = 5\sqrt{3}\frac{1}{2} = 2\sqrt{3}$$.

The y component of initial velocity of projectile 1. $$u_{y}= u sin 60^{0} = 5 \sqrt{3}\frac{\sqrt{3}}{2} = 7.5$$.

and $$x_{0} = 2.5 \sqrt{3}t_{0}$$

also $$y_{0} = 10 + 7.5 t_{0} - \frac{1}{2}gt_{0}^{2}$$

After the second projectile is in air and on a collision course, both have the same acceleration g in vertically downward direction. So, their relative acceleration is zero i.e. they move with constant velocity with respect to each other. We place an observer with projectile 1 at time t0, for which the second projectile is coming to hit it in a time interval T with a constant velocity, for which the second projectile will have following displacements to cover

Along x -axis = $$2.5\sqrt{3}t_{0}$$

and along y - axis = $$10 - y_{0} = -7.5 t_{0}+\frac{1}{2}gt^{2}$$

The velocity components available to the second projectile to cover up the respective displacements are

x - component of $$v_{rel} = 5\sqrt{3} - 2.5\sqrt{3} = 2.5\sqrt{3}$$

y - component of $$v_{rel} = 7.5 - 10t_{0}$$

The equation of motion of projectile 2 with respect to projectile 1 are as following

Along x axis

$$x_{0} = 2.5\sqrt{3} T$$

This gives $$t_{0} =T$$

Along y axis

$$- 7.5t_{0} + 5 t_{0}^{2} = (7.5 - 10 t_{0})T$$, Putting $$t_{0} =T$$ and solving we get $$t_{0} = 1$$ s. Accordingly , Coordinates of point $$P = (5\sqrt{3},5)$$