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November 26, 2014

November 26, 2014

Posted by **Alex** on Sunday, October 30, 2011 at 5:30pm.

(a) How long does the ball take to reach the wall?

(b) How far above the release point does the ball hit the wall?

(c) What are the horizontal and vertical components of its velocity as it hits the wall?

- Physics -
**Henry**, Tuesday, November 1, 2011 at 5:37pmVo = 30m/s @ 40deg.

Xo = hor. = 30cos40 = 23m/s.

Yo = ver. = 30sin40 = 19.28m/s.

a. Dh = VT = Hor. dist.

T = Dh / V = 22 / 23 = 0.96s =.

2t = T = 0.96,

t = 0.96 / 2 = 0.48s to reach wall.

b. t(up) = T/2 = 0.96 / 2 = 0.48s.

d(up) = Vo*t + 4.9t^2,

d(up)=19.28*0.48 -4.9(0.48)^2 = 8.125m

c. X = Xo = hor = 30cos40 = 23m/s.

Y^2 = Yo^2 + 2gd,

Y^2 = (19.28)^2 - 19.6*8.125 = 212.47,

Y = ver. = 14.6m/s.

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