A stone A is dropped from the top of a very high platform. At the same time another stone is thrown vertically upward from the ground level with an initial speed of 50m/s. At the instance that the stone passes each other, the ration of the velocities of A and B is 2:3 respectively. Evaluate (i) time taken for stones to pass each other (ii) height of the platform (iii) height of above the ground when the velocities are equal and opposite (iv) maximum height reached by B (v) time interval between the arrival of A and B at the ground

Answer 1

va = gt
vb = 50 - gt
so
gt/(50-gt)) = 2/3
solve for t
ii) Now use this time to figure how far a has fallen (1/2gt^2) and how far b went up (50t-1/2 g t^2). Add for total height
iii) Same as a except make the velocities equal.
iv) 50^2 = 2 g y. Solve for y.
v) Time for a to fall height in ii vs time for b to land again (y=0)