You are standing on the top of a building 135 meters tall. You throw a ball upward with a velocity of 22.0 m/s.
At the exact same moment a friend throws a second ball upward from the ground with a velocity of 46.0 m/s. these two
balls then collide at some later time. How long after these two balls are released will they collide?

Now i know the answer is 5.63 seconds since the teacher gave us the answer
but what he is looking for is to show our work by writing the information that we are given,
finding the formula and plugin in the numbers to work it out, please help.

Question

You are standing on the top of a building 135 meters tall. You throw a ball upward with a velocity of 22.0 m/s.
At the exact same moment a friend throws a second ball upward from the ground with a velocity of 46.0 m/s. these two 
balls then collide at some later time. How long after these two balls are released will they collide?

Now i know the answer is 5.63 seconds since the teacher gave us the answer
but what he is looking for is to show our work by writing the information that we are given,
finding the formula and plugin in the numbers to work it out, please help.

anonymous · Answer

T = 135/(46-22)
   = 135/24
   = 5.625=5.63sec
SOLUTION: Time at which they collide/meet is T=H/U
H is height of the building
U is the initial velocity of the ball
here, U1 and U2 are given for two different balls
So, T=H/(U2-U1)
here we subtract the initial velocities as the balls move in opposite direction to each other
U2 is 46
U1 is 22
H is 135
substitute the above values in T=H/(U2-U1)

hope this  helps u...
cheers!!