# A ball is thrown upward from the ground with an initial speed of 25ms^-1. At the same instant , another ball is dropped from a 15m building . When wil two balls be at the same height from the ground?

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We can use the equation of motion

h = ut + 0.5at^2

where, h is height, t is time, u is initial velocity and a is acceleration (g for objects falling down and -g for objects thrown upwards).

For the ball thrown upwards,

h = 25t - 0.5 gt^2

or, 25t - h = 0.5gt^2

And for ball dropped downwards,

15-h = 0.5gt^2

Comparing the two equations, we get

25t-h = 15-h

Or, t = 0.6 sec.

You can calculate using this value of time and find that both the balls will be at 13.24 m from surface.

Let h be common height of both the balls after time t.

Now,

For the ball which is thrown upwards,we use the formula

`h = 25t - (g/2)t^2`

=> `25t-h = (g/2)t^2`

and for the ball dropped from the height of 15 m,

`15 - h = (g/2)t^2`

Solving the two equations we get the value of t

`15 - h = 25t - h`

=> `t= 15/25` sec.

= `0.6 ` sec.