Two stones are thrown up simultaneously from the edge of cliff with initial speeds v and 2v . The relative position of the second stone with respect to first varies with time till both the stones strike the ground as

Question

Two stones are thrown up simultaneously from the edge of cliff with initial speeds v and 2v . The relative position of the second stone with respect to first varies with time till both the stones strike the ground as (A) Linearly (B) First linearly then parabolically (C) Parabolically (D) First parabolically then linearly

Tardigrade Admin · Accepted Answer

Initially when both are in air, the relative acceleration between them is zero. Therefore, relative position verus-time graph is a straight line.

S_{1} = v t - \frac{1}{2} g t^{2}

S_{2} = 2 v t - \frac{1}{2} g t^{2}

S_{21} = S_{2} - S_{1}

S_{21} = v t

Once the first ball reaches the ground, its acceleration becomes zero. Therefore, the relative acceleration is now g and the graph is parabolic.

S_{2} = 2 v t - \frac{1}{2} g t^{2}

Q. Two stones are thrown up simultaneously from the edge of cliff with initial speeds v and 2v . The relative position of the second stone with respect to first varies with time till both the stones strike the ground as

Linearly

First linearly then parabolically

Parabolically

First parabolically then linearly

Q. Two stones are thrown up simultaneously from the edge of cliff with initial speeds $v$ and $2 v$ . The relative position of the second stone with respect to first varies with time till both the stones strike the ground as