Question

A body is dropped from the top of the tower 400mhigh. At the same time another body is projected vertically occurred which velocity is 100m/s. Find when and where they will meet​

Answer 1

Answer:

80 m from above and time required is 4 secs.

Explanation:

Let x be a distance from top where they will meet.

x =(1/2)g(t^2)

Also ,

400-x = 100t-(1/2)g(t^2)

Now adding both equations

400=100t

Therefore t =4secs

And x=80 m