A skateboarder shoots off a ramp with a velocity of 7.5 m/s,directed at an angle

Question

A skateboarder shoots off a ramp with a velocity of 7.5 m/s,directed at an angle of 54° above the horizontal. The end ofthe ramp is 1.4 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 the ramp?

Explanation / Answer

We have these equation x=vcost y=y0+vsint-gt^2/2 when the skateboarder reach the highest point, its verticalvelocity is zero. so t'=vsin/g=0,619(s). y=1,4+vsint-gt^2/2=3,28(m) x=vcost=2,73(m)

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