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

You throw a ball with a speed of 30 m/s at an angle of 40.0° above the horizontal directly toward a wall. The wall is 22.0 m from the release point of the ball.

(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 5:37pm
Vo = 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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