posted by Emily .
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
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
thanks a lot@Damon