Two guns, situated of the top of a hill of height10 m, fire one shot each with the same speed 5 root3 m/s at some interval of time. One gun fireshorizontally and other fire upwards at an angleof 60º with the horizontal. The shots collide inair at a point P. Find (I.I.T. 1996)

i) The time interval between the firings, and

(ii) The coordinates of the point P. Take origin of the coordinates system at the footof the hill right below the muzzle andtrajectories in X-Y plane.

Mohammed Patel, 10 years ago

Grade:12

1 Answers

himanshu porwal

29 Points

10 years ago

Shot A - The one which was launched at an angle. Shot B - The one which was launched horizontally.

This is the first point which you need to think about. Which one of the two was launched first? Think about it, they have to collide. A has to cover the same horizontal displacement as B, but with a lesser horizontal velocity as launched at an angle.

If they collide, it just means that they are at the same position at the same instant of time. The two projectiles start from the same point, which means, in order to collide - their displacements must be same at a particular time instant.

For horizontal displacement, we have [Hor. velocity of Particle 1] x [Time till collision] = [Hor. Vel. of particle 2] x[Time till collision] eqn.(1) .

The horizontal motion of projectiles is uniform.

All we have to do now is to substitute the values. Note that the time till collision for the two particles is not the same! If for the second particle it is T then for the first particle, it is T+t where t is the time interval between the firings.

Sx = Vxa x (T+t) = VxB x (T) Vxa=velocity of A horizontally,Vxb=velocity of B horizontally

Sy = VyA x (T+t) - (0.5)(g)(T+t)^2 = VyB x (T) - (0.5)(g)(T)^2.

This equation gives us T=t, after substituting the two velocities.

Substituting T=t from the first equation and solving, we get t=1 second interval between the firings.

For the second part, we require horizontal and vertical displacements. Go back to the above equations, plug in t=1/T=1. We get Sx = 5(3)^0.5 and Sy = -5 m

The displacemtent is the final-initial coordinate, so X = 5root3 m, Y = -5 + 10 = 5 m. [Launched from (0,10),]