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A ball is thrown straight up with a speed of 21.5 m/s. A second ball is thrown up 1.52 seconds later and hits the first ball just the first ball reaches its maximum height. What was the magnitude of the velocity of the second ball?

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Added Wed, 06 Apr '16
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first ball

t1 = v1/g = 21.4/9.8 = 2.18 s

t2 = 2.18 - 1.52 = 0.66 s

h = v1^2/2g = 23.26 m

for second ball

h = v2*t2 - 0.5*g*t2^2

23.26 = v2*0.66 - 0.5*9.8*0.66^2

v2 = 38.47 m/s

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Added Wed, 06 Apr '16
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