Posted by Ryan on Saturday, December 3, 2011 at 3:30pm.
A ball is thrown straight upward and returns
to the thrower’s hand after 2.90 s in the air.
A second ball is thrown at an angle of 32.0
◦
with the horizontal.
At what speed must the second ball be
thrown so that it reaches the same height as
the one thrown vertically? The acceleration
of gravity is 9.81 m/s
2
.
Answer in units of m/

Physics  Henry, Sunday, December 4, 2011 at 11:13am
First Ball:
Tr = Tf = T/2 = 2.9/2 = 1.45s = Time to
rise to max. ht.
Vf = Vo + gt,
Vo = Vf  gt,
Vo = o + 9.81*1.45 = 14.22m/s.=Initial velocity of 1st ball.
h = Vo*t + 0.5g*t^2,
h = 14.22*1.45  4.9*(1.45)2 = 10.32m.
Second Ball:
Yo = Vo*sin32 = 14.22m/s,
Vo = 14.22 / sin32 = 26.83m/s @ 32 deg = Inital velocity of 2nd ball.
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