A skateboarder shoots off a ramp with a velocity of 6.1 m/s, directed at an angl
ID: 2046482 • Letter: A
Question
A skateboarder shoots off a ramp with a velocity of 6.1 m/s, directed at an angle of 58degrees 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. When the skateboarder reaches the highest point, how far is this point horizontally from the end of the ramp?Explanation / Answer
initial vertical velocity of the skateboarder=5.8 sin54 vertical acceleration = g = 9.8m/s2 let the skateboarder reach a height h above the end ofthe ramp u =5.8*sin54 v = 0 using basic kinematic equation v2 = u2-2*g*h 0 =(5.8*sin54)2-2*9.8*h ==> h = 1.123 m b) the highest point is 1.3 +1.123=2.423m above the ground time taken to reach thehighest point=t v =u-g*t 0 =5.8*sin54 - 9.8*t t =0.478 s initial horizontalvelocity=5.8*cos54 horizontal acceleration =0 distance traveled in 0.478 s horizontally =5.8*cos54*0.478 = 1.629 m the highest point is 1.629 m from theend of the ramp