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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,

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,


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Steve Johnson


I would be delighted to help with questions up through the first year of college Physics. Particularly Electricity, Electronics and Newtonian Mechanics (motion, acceleration etc.). I decline questions on relativity and Atomic Physics. I also could discuss the Space Shuttle and space flight in general.


I have a BS in Physics and an MS in Electrical Engineering. I am retired now. My professional career was in Electrical Engineering with considerable time spent working with accelerometers, gyroscopes and flight dynamics (Physics related topics) while working on the Space Shuttle. I gave formal classroom lessons to technical co-workers periodically over a several year period.

BS Physics, North Dakota State University
MS Electrical Engineering, North Dakota State University

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