# Physics HELP!!!

A ball is kicked at an angle of 50 with a velocity of 14m/s.
a. how long is it in the air?
b. how high does it rise?

can someone please explain to me how to solve this???

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1. horizontal velocity, U = 14 cos 50 = 9
Since there is no horizontal force on the ball, the horizontal velocity will remain 9.

Initial vertical velocity, Vo = 14 sin 50 = 10.7

vertical velocity at the top = 0 (it stops and starts down)
v = Vo - 9.8 t
0 = 10.7 - 9.8 t
t = 1.09 seconds to the top
It will spend the same time falling so a total of 2.18 seconds in the air.

now how high at 1.09 seconds?
h = 0 + Vo t -(1/2)9.8 t^2
h = 10.7 (1.09) - 4.9 * 1.19
h = 5.83 meters (about 17 feet)

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posted by Damon
2. thanks! I have question though...what happened to the ^2 in the second part? Its in the second step but then where does it go?

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posted by Lauren
3. nevermind, i figured it out

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posted by Lauren

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