From the same point, two stones A and B are thrown simultaneously, A is thrown up vertically with a velocity of 10 ms⁻¹ and B is thrown up with the same velocity at 30^ with the horizontal. The separation between the balls at 2 s is

Step-by-step solution

Let the point of projection be the origin (0,0) . Position vector of stone A at time t is: r _A = (10t - 1 2 gt^2) j Position vector of stone B at time t is: r _B = (10 30^ t) i + (10 30^ t - 1 2 gt^2) j r _B = 5 3 t i + (5t - 1 2 gt^2) j The relative position vector of A with respect to B is: r _ AB = r _A - r _B = -5 3 t i + 5t j The separation between the stones is the magnitude of r _ AB : | r _ AB | = (-5 3 t)^2 + (5t)^2 | r _ AB | = 75t^2 + 25t^2 = 100t^2 = 10t At t = 2 s , the separation is: d = 10 2 = 20 m