The second ball hits when
60-4.9(t-2)^2 = 0,
t = 5.5
v = 19.6(1 - 1/5.5) = 10.036 m/s
So, you want v where
60 + 5.5v - 4.9*5.5^2 = 0
v = 16.04 m/s
check:
A ball is thrown straight up from the edge of the roof of a 60.0 m tall building. If a second ball is
dropped from the roof 2.0 s later what must the initial velocity of the first ball be if both are to reach
the ground at the same time?
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