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A ball is thrown vertically upwards with an initial velocity such that it can reach a maximum height of 15 m.If,at the same instance ,a stone is dropped from a height of 15 m,find the ratio of distances travelled by them when they cross each other.

Hello Abhishek,

There are 2 facts about this problem that we can take advantage of. Let's consider up to be the positive direction. When they cross, the ball is at height h and the stone will have fallen a distance -15 m + h. And the expired times since the ball was thrown and since the stone was dropped is t, for both of them. These facts will help us set up and solve simultaneous equations.

Also, we can determine the initial speed of the ball from the fact that it can reach a height of 15 m. Use the kinematic formula

Vf^2 = Vi^2 + 2*a*y

where Vf = 0, a = -g = -9.8 m/s^2, and y = 15 m. Solve for Vi.

Now, we need to write those simultaneous equations.

The ball will reach height h in time t.

h = Vi*t + (1/2)*a*t^2

Plug in your value of Vi and the value of a used above. Call that equation 1.

The stone will fall a distance of -15 m + h in time t.

-15 m + h = Vi*t + (1/2)*a*t^2

where Vi for the stone is zero, and a is the same value used before. Plug in that data and call that equation 2.

Plug the expression for h in equation 1 in for the h in equation 2. Solve for t. Plug that value of t into equation 1 to find h.

Then, technically, your answer should be h/(-15 m + h). But I would bet they don't want a negative number here. So drop the minus sign.

I hope this helps,

Steve

Physics

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