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October 25, 2014

October 25, 2014

Posted by **Esora** on Sunday, October 31, 2010 at 9:38pm.

-What is the maximum height reached by the ball?

-What is the horizontal distance traveled by the ball?

- physics -
**Damon**, Sunday, October 31, 2010 at 9:47pmVi = 30 sin 38

h = Ho + Vi t - 4.9 t^2

h = 0 at ground

so

0 = 0 + Vi t - 4.9 t^2

t (Vi-4.9 t) = 0

t = 0 (it started at ground

t = Vi/4.9 when it hit ground again

max h is when v = 0

v = Vi -9.8 t

0 = Vi -9.8 t

t = Vi/9.8 at top *** this should be half the whole time in the air***

h = 0 +Vi t - 4.9 t^2

using that t which should be half the total t

u = 30 cos 38

d = u t where t is the total time in the air (not the half time you used for part B)

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