A ball is thrown up at initial speed of 25 m/s at the same time a ball is dropped from a building 14 m high. when will they both be the same height?

One ball is thrown vertically at 25 m/s.

The other ball is let go at a height of 14 m, at a speed of 0 m/s.

We need to realize that gravity will increase the speed of the ball that is dropped and gravity will decrease the speed of the ball that goes up.

Imagine the balls meet at a height h.

Then the ball that is thrown up travels:

h=vot-1/2at^2

h=25t-1/2gt^2 EQN 1

During this same amount of time, t,

the ball falling will travel:

h-15=vot+1/2at^2

h-15=0+1/2(-g)^2

h=15-1/2gt^2 EQN 2

Substitute for h:

25t-1/2gt^2=15-1/2gt^2

25t=15

t=0.6 seconds

This is the correct answer. I guarantee it.

I can solve many more problems like this. Let me know when you need help for these types of problems.

Secure Information

Content will be erased after question is completed.

Enter the email address associated with your account, and we will email you a link to reset your password.

Forgot your password?

Sign Up