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