A skateboarder shoots off a ramp with a velocity of 5.9 m/s, directed at an angl
ID: 249913 • Letter: A
Question
A skateboarder shoots off a ramp with a velocity of 5.9 m/s, directed at an angle of 56° above the horizontal. The end of the ramp is 1.1 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?
Explanation / Answer
U = 5.9 m/s at 56°
Ux = 5.9cos56 = 3.3m/s
Uy = 5.9sin56 = 4.89m/s
g = 9.8m/s²
a)
max height is when Vy = 0 so
0 = Vy² = Uy² - 2gy
gives the max height as
y = Uy²/(2g) = 4.89²/19.6 = 1.22m
b) also if Vy=0 then
0 = Vy = Uy - gt
so the time is takes to reach max height is
t = Uy/g
then the horizontal distance is
x = Uxt = UxUy/g = 3.3 * 4.89 / 9.8 = 1.65m