Posted by Jack on .
A skateboarder shoots off a ramp with a velocity of 6.8 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?
upward component of velocity at takeoff is 6.8 sin 56° = 5.64
h(t) = 1.1 + 5.64t - 4.9t^2
max height reached at t = 5.64/9.8 = 0.5755
h(0.5755) = 2.723