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

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)

To solve this problem, we can use equations of motion and kinematic principles. First, let's break down the problem into parts:

(i) Time taken for stones to pass each other:
Let's assume that stone A takes time t to reach the ground from the top of the platform. At the same time, stone B takes time t to reach the highest point in its trajectory and come back down. The time taken for stones A and B to pass each other is the sum of their respective times of flight, which is t + t = 2t.

(ii) Height of the platform:
Since stone A is dropped from the top of the platform, its initial velocity is zero. The distance fallen by stone A can be calculated using the equation:

h = (1/2)gt^2

where h is the height, g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time of flight. In this case, h represents the height of the platform.

(iii) Height above the ground when the velocities are equal and opposite:
When the velocities of the stones are equal and opposite, stone A and stone B have the same magnitude of velocity but in opposite directions. This occurs when stone A is at its maximum velocity while falling, and stone B is at its highest point. At this point, both stones have the same gravitational potential energy.

Using the equation for gravitational potential energy:

mgh = (1/2)mv^2

where m is the mass of the stone, g is acceleration due to gravity, h is the height above the ground, and v is the velocity. Since the masses of the stones are not given and they will cancel out in the ratio calculations, we don't need to consider mass. Let's call the height above the ground at this point as h'.

(iv) Maximum height reached by B:
The maximum height reached by stone B can be calculated using the equation for vertical motion:

v^2 = u^2 - 2gh

where v is the final velocity (which is zero at the highest point), u is the initial velocity (50 m/s), g is acceleration due to gravity, and h is the maximum height. Rearranging the equation, we get:

h = (u^2)/(2g)

Substituting the given values, we can calculate the maximum height reached by stone B.

(v) Time interval between the arrival of A and B at the ground:
The total time taken by each stone (A and B) to hit the ground from their respective starting positions can be calculated using the equation for vertical motion:

h = ut + (1/2)gt^2

Solving this equation for time t, we can calculate the time taken by each stone to reach the ground. The time interval between the arrival of stones A and B at the ground is the difference between their respective times of flight.

So, to summarize:
(i) The time taken for stones A and B to pass each other is 2t.
(ii) The height of the platform is h.
(iii) The height above the ground when the velocities are equal and opposite is h'.
(iv) The maximum height reached by stone B is h.
(v) The time interval between the arrival of stones A and B at the ground is the difference between their respective times of flight.