A ball A is thrown vertically upward from the top of a 30-m-high building with an initial velocity of 5 m>s. At the same instant another ball B is thrown upward from the ground with an initial velocity of 20 m>s. Determine the height from the ground and the time at which they pass.

s = 20.4 m

Explanation:

First lets write down equations for each ball:

s=so+vo*t+1/2a_c*t^2

for ball A:

s_a=30+5*t+1/2*9.81*t^2

for ball B:

s_b=20*t-1/2*9.81*t^2

to find time deeded to pass we just put that

s_a = s_b

30+5*t-4.91*t^2=20*t-4.9*t^2

t=2 s

now we just have to put that time in any of those equations an get distance from the ground:

s = 30 + 5*2 - 1/2*9.81 * 2^2

s = 20.4 m