2 Answers | Add Yours
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.
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.
We’ve answered 319,655 questions. We can answer yours, too.Ask a question