A skateboarder shoots off a ramp with a velocity of 6.5 m/s, directed at an angl
ID: 1955436 • Letter: A
Question
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 1.1 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
To get the highest point, you know the vertical velocity at that point is 0 m/s. To get the initial vertical velocity: sin(angle) = Vy / V sin(63 degrees) = Vy / 6.5 Vy = 5.79 m/s To get the highest point, you can now use: Vf^2 = Vi^2 + 2 * a * d 0 = 5.79^2 + 2 * (-9.8) * d d = 1.71 meters So total height = 1.71 meters + 1.1 meters = 2.81 meters is the max height. To find the horizontal distance, you need the horizontal velocity: cos(angle) = Vx / V cos(63 degrees) = Vx / 6.5 Vx = 2.95 m/s Now to find the distance traveled, you need time as well: Vf = Vi + a*t 0 = 5.79 + (-9.8) * t t = 0.59 seconds So total horizontal distance is: D = V*t D = 2.95 * 0.59 D = 1.74 meters