A skateboarder shoots off a ramp with a velocity of 7.1 m/s,directed at an angle

Question

A skateboarder shoots off a ramp with a velocity of 7.1 m/s,directed at an angle of 56° above the horizontal. The end ofthe ramp is 1.5 m above the ground. Let the x axis beparallel to the ground, the +y direction be verticallyupward, and take as the origin the point on the ground directlybelow the top of the ramp. (a) How high above theground is the highest point that the skateboarder reaches?(b) When the skateboarder reaches the highestpoint, how far is this point horizontally from the end of theramp?

Explanation / Answer

We know that
             v2 - u2 = 2*a*h ==>       0 -(5.886)2 = - 2*9.8*h Therefore the high above the ground is the highest point thatthe skateboarder reaches is                       = 1.5 + 1.767
                      = 3.267 m
The time taken to reach the highest point is
                     t = u / g
                       = 5.886 / 9.8
                       = 0.6006 s
(b)
Initial horizontal velocity u ' = 7.1 *cos56 = 3.970m/s
horizontal acceleration = 0 Therefore the distance traveled in 0.6006 s is
d= u' * t
                    = 3.970 * 0.6006
                    = 2.384 m

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