Question

A skateboarder shoots off a ramp with a velocity of 5.1 m/s, directed at an angle of 55° above the horizontal. The end of the ramp is 1.0 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?

Answer 1

Answer

given,

initial velocity of skateboard = 5.1 m/s

angle above the horizontal = 55°

height of the ramp = 1 m

a) maximum height of projectile

  $H = \dfrac{u^2sin^2 \theta}{2g}$

  $H = \dfrac{5.1^2\times sin^2 55^0}{2\times 9.81}$

         H =  0.889 m

the maximum height of the skateboard above the ground

         = 1 + 0.889

         = 1.889 m

b) time to reach the height

   $t = \dfrac{u\ sin\theta}{g}$

   $t = \dfrac{5.1\ sin55^0}{9.8}$

          t = 0.426 s

horizontal distance = u cos θ × t

                                = 5.1 × cos 55° × 0.426

horizontal distance = 1.25 m