Question

Corbin, an avid skier leaves the horizontal end of a ramp with a velocity of 33.0 m/s and lands 63.0 m from the base the ramp. How high is the end of the ramp from the ground?

Answer 1

Answer:

 Height of ramp = 17.49 m

Explanation:

  We have equation of motion , $s= ut+\frac{1}{2} at^2$, s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

 Considering horizontal motion of skier

      Initial velocity = 33 m/s, Displacement = 63 m, acceleration = 0 and we need to find time taken to reach ground by the skier.

    $63= 33t+\frac{1}{2} *0*t^2\\ \\ t=1.909seconds$

The vertical distance traveled in 1.909 seconds is the height of  ramp

      Initial velocity = 0 m/s, acceleration = acceleration due to gravity =  9.8 $m/s^2$, time = 1.909 s and we need to find displacement.

      $s= 0*1.909+\frac{1}{2} *9.8*1.909^2\\ \\ s=17.49 m$

    So height of ramp = 17.49 m