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September 18, 2014

September 18, 2014

Posted by **Justin** on Tuesday, June 18, 2013 at 9:44pm.

- Physics -
**Justin**, Tuesday, June 18, 2013 at 10:09pm1. a) Dy = Viy(t) + (1/2)(g)t^2

15.5 m = 0(t) + (1/2)(9.8 m/s^2)(t^2)

t = sqrt[(15.5 m)/(4.9 m/s^2)]

t = 1.8 sec answer

b) Let V = speed of the ball just before it strikes the water.

Vix = Vfx = 11.4 m/s

Vfy = Viy + gt

Vfy = 0 + (9.8 m/s^2)(1.8 s)

Vfy = 17.64 m/s

V = sqrt[Vfx^2 + Vfy^2]

V = sqrt[(11.4 m/s)^2 + (17.64 m/s)^2]

V = 21.0 m/s

It works I just used different numbers to test.

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