A skateboarder shoots off a ramp with a velocity of 5.8 m/s,directed at an angle

Question

A skateboarder shoots off a ramp with a velocity of 5.8 m/s,directed at an angle of 54° above the horizontal. The end ofthe ramp is 1.3 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

vertical acceleration = g = 9.8m/s2
let the skateboarder reach a height h above the end ofthe ramp    u =5.8*sin54    v = 0 using basic kinematic equation
    v2 = u2-2*g*h      0 =(5.8*sin54)2-2*9.8*h     ==> h = 1.123 m
b)
     the highest point is 1.3 +1.123=2.423m above the ground
       time taken to reach thehighest point=t          v =u-g*t           0 =5.8*sin54 - 9.8*t           t =0.478 s
      initial horizontalvelocity=5.8*cos54       horizontal acceleration =0     distance traveled in 0.478 s horizontally =5.8*cos54*0.478 = 1.629 m

     the highest point is 1.629 m from theend of the ramp

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