posted by Anonymous .
A skateboarder shoots off a ramp with a velocity of 6.5 m/s, directed at an angle of 63° above the horizontal. The end of the ramp is 1.5 m above the ground. Let the x axis be parallel to the ground, the +y direction be vertically upward, and take as the origin the point on the ground directly below the top of the ramp.
(a) How high above the ground is the highest point that the skateboarder reaches?
(b) When the skateboarder reaches the highest point, how far is this point horizontally from the end of the ramp?
(a) The vertical rise H can be determined by requiring that the vertical portion of the kinetic energy, (1/2)M*(V sin 63)^2, be converted to potential energy M g H.
H = (1/2)(V sin63)^2/g
Add H to the initial platform height, 1.5 m, for the highest point elevation above ground.
(b) Multiply the time to reach maximum height (zero vetical velocity), (Vsin63)/g, by the horizontal velocity component, V cos63.
X = (V^2/g)sin63 cos63