A skateboarder shoots off a ramp with a velocity of 7.3 m/s, directed at an angl
ID: 1954891 • Letter: A
Question
A skateboarder shoots off a ramp with a velocity of 7.3 m/s, directed at an angle of 58° 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
Consider the vertical component of the skateboarder's initial velocity when he leaves the ramp, 7.3 sin 58° m/s. Applying equations of motion in the +y direction, we know
u = 7.3 sin 58° m/s
v = 0 m/s at the top of his trajectory
a = -9.8 m/s2
To solve for s
The equation that involves these four variables is v2 = u2 + 2as. Solve for s, the vertical displacement, and remember to add 1.4 m to account for the fact that the top of the ramp is 1.4 m from the ground.
To solve for t (we will see why later), use v = u + at.
To know the horizontal distance of the highest point of the skateboarder's trajectory from the end of the ramp, apply, in the x direction, s = ut. u refers to the initial horizontal velocity, 7.3 cos 58° m/s. Solve for s, the horizontal displacement.