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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