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# 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:

$s=s_0+v_0t+\frac{1}{2}a_ct^2$

for ball A:

$s_A=30+5t-\frac{1}{2}\cdot9.81t^2$

for ball B:

$s_B=20t-\frac{1}{2}\cdot9.81t^2$

to find time deeded to pass we just put that

$s_A=s_B$

:

$30+5t-4.91t^2=20t-4.91t^2$

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

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

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

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

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