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Question

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.

Solution

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First lets write down equations for each ball:

s=s0+v0t+12act2s=s_0+v_0t+\frac{1}{2}a_ct^2

for ball A:

sA=30+5t129.81t2s_A=30+5t-\frac{1}{2}\cdot9.81t^2

for ball B:

sB=20t129.81t2s_B=20t-\frac{1}{2}\cdot9.81t^2

to find time deeded to pass we just put that

sA=sBs_A=s_B

:

30+5t4.91t2=20t4.91t230+5t-4.91t^2=20t-4.91t^2

t=2s\boxed{t=2\,\,\rm s}

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

s=30+52129.8122s=30+5\cdot2-\frac{1}{2}\cdot9.81\cdot2^2

s=20.4m\boxed{s=20.4\,\,\rm m}

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