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A ball is dropped from 160m high tower and at the same time, a stone is projected vertically upwards with 40m/s. Find the distance from ground when stone will meet.

Please answer it fast sir,

thanks

Hello Ankit,

There are a few useful things we can say at the start. The elapsed time, t, is the same for both ball and stone. If the altitude when they meet is y, the ball drops a distance of 160 m - y. Let us choose up to be the positive direction. Therefore the ball's displacement is y - 160 m. Now we can plug our data into the kinematic formula

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

For the stone:

y = 40 m/s*t + (1/2)*(-9.8 m/s^2)*t^2

For the ball:

y - 160 m = (1/2)*(-9.8 m/s^2)*t^2

Solving the last equation for y

y = (1/2)*(-9.8 m/s^2)*t^2 + 160 m

Now we can set the 2 equations for y equal to each other.

40 m/s*t + (1/2)*(-9.8 m/s^2)*t^2 = (1/2)*(-9.8 m/s^2)*t^2 + 160 m

40 m/s*t = 160 m

t = 160 m / (40 m/s) = 4 s

So now you can use that value of time in either of the first 2 equations and find the value of y. I will leave that for you to complete.

I hope this helps,

Steve

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