Posted by **Emily** on Sunday, January 30, 2011 at 1:35pm.

For a science fair competition, a group of High School students built a kicker-machine that can launch a golf ball from the origin with a velocity of 12.3 m/s and initial angle of 29.9° with respect to the horizontal.

*Where will the golf ball fall back to the ground?

*How High will it be at it's highest point of trajectory?

*What is the ball's velocity vector (in Cartesian components) at the highest point of its trajectory? x, then y

*What is the ball's acceleration vector (in Cartesian components) at the highest point of its trajectory? x, then y

- Physics -
**Damon**, Sunday, January 30, 2011 at 1:45pm
Answer the last two questions immediately.

At the highest point the ball has no y velocity and the x velocity is the same for the whole trip, u = 12.3 cos 29.9

During the whole trip the acceleration is only in the y direction and is -9.8m/s^2

Now do the problem.

Y(vertical) problem first.

Vi = 12.3 sin 29.9

at top v = 0

0 = Vi - 9.8 t

9.8 t = 12.3 sin 29.9

solve for t at top, which is half the trip time

h = Vi t - 4.9 t^2

solve for h, height at top

T = 2 t, total time in air

then d = u T = [12.3 cos 29.9 ] * T

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