A skateboarder shoots off a ramp with a velocity of 6.0 m/s, directed at an angl
ID: 1538953 • Letter: A
Question
A skateboarder shoots off a ramp with a velocity of 6.0 m/s, directed at an angle of 62 degree above the horizontal. The end of amp is 1.4 m above the ground. Let the x axis be parallel to the ground, the +y direction be vertically upward, and take the origin the point on the ground directly below the top of the ramp, How high above the ground is the highest point that the skateboarder reaches? When the skateboarder reaches the highest point, how far is this point horizontally from the end of the ramp? By accessing this Question Assistance, you will learn while you earn points based on the Point Potential Policy set by your instructor.Explanation / Answer
a) Y = Uyt - 0.5gt^2
-1.4 = 6 sin 62 t - 0.5*9.8*t^2
4.9t^2 - 5.3t - 1.4 = 0
t = 5.3 + sqrt(5.3^2 +4*4.9*1.4) / 9.8
= 1.3 s
a) Vy^2 - Uy^2 = 2g*(Hmax)
0 - (6 sin 62)^2 = 2*9.8*(Hmax)
(Hmax) = 1.43 m
so height above ground = 1.43+1.4 = 2.83 m
b) X = Ux*t
it will reach the maximum time at half the time of flight
X = 6 cos 62 * (1.3/2)
= 1.83 m