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.

## We’ll help your grades soar

Start your 48-hour free trial and unlock all the summaries, Q&A, and analyses you need to get better grades now.

- 30,000+ book summaries
- 20% study tools discount
- Ad-free content
- PDF downloads
- 300,000+ answers
- 5-star customer support

Already a member? Log in here.

Are you a teacher? Sign up now