Askateboarder shoots off a ramp with a velocity of 6.6 m/s, directed at an angle of 58 above the horizontal. the end of the ramp is 1.2 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? (b) when the skateboarder reaches the highest point, how far is this point horizontally from the end of the ramp?

a) Maximum height reached above ground = 2.8 m

b) When he reaches maximum height he is 2 m far from end of the ramp.

Explanation:

a) We have equation of motion v²=u²+2as

   Considering vertical motion of skateboarder.  

   When he reaches maximum height,

          u = 6.6sin58 = 5.6 m/s

          a = -9.81 m/s²

          v = 0 m/s

  Substituting

         0²=5.6² + 2 x -9.81 x s

          s = 1.60 m

  Height above ground = 1.2 + 1.6 = 2.8 m

b) We have equation of motion v= u+at

   Considering vertical motion of skateboarder.  

   When he reaches maximum height,

          u = 6.6sin58 = 5.6 m/s

          a = -9.81 m/s²

          v = 0 m/s

  Substituting

         0= 5.6 - 9.81 x t

          t = 0.57s

  Now considering horizontal motion of skateboarder.  

  We have equation of motion s =ut + 0.5 at²

          u = 6.6cos58 = 3.50 m/s

          a = 0 m/s²

          t = 0.57  

  Substituting

         s =3.5 x 0.57 + 0.5 x 0 x 0.57²

         s = 2 m

  When he reaches maximum height he is 2 m far from end of the ramp.

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