posted by Ryan .
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
Answer in units of m/
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.
Yo = Vo*sin32 = 14.22m/s,
Vo = 14.22 / sin32 = 26.83m/s @ 32 deg = Inital velocity of 2nd ball.