A skateboarder shoots off a ramp with a velocity of 6.4 m/s, directed at an angl
ID: 2020257 • Letter: A
Question
A skateboarder shoots off a ramp with a velocity of 6.4 m/s, directed at an angle of 56° above the horizontal. The end of the ramp is 1.4 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
to find the height above the ramp use the equation; (vf^2 -voy^2)/2g = h, with vf at top equal to 0, voy = vosin (theta) = 6.4m/s sin (56)=5.3m/s, and g = -9.8 because it was upward. plugging those values in the above equation gives h= 1.44m, but the question asks above the ground. so, answer is 1.44m+1.4m = 2.84m. you can also use y = voy t -1/2 gt^2. b) time to reach top equals (vf-voy)/g, where vf = 0, voy = 5.3m/s,g = 9.8m/s^2. that gives you t= 0.54s horizontal distance = vox*t, where vox = vo cos(theta) = 6.4cos(56)=3.58m/s hence horizontal distance = 3.58m/s * 0.54s = 1.93meters.