Question

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?

Accepted Answer

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.

Answer

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.

Science

A Ball Is Thrown Upward From The Ground

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