Question

After leaving the end of a ski ramp, a ski jumper lands downhill at a point that is displaced 36.0 m horizontally from the end of the ramp. His velocity, just before landing, is 19.0 m/s and points in a direction 33.2 ° below the horizontal. Neglecting air resistance and any lift that he experiences while airborne, find (a) the magnitude and (b) the direction of his initial velocity when he left the end of the ramp. (a) Number Units

Answer 1

Answer

v = 18.99 m/s

$\theta = 33.20^0$

Explanation:

given,

displacement = 36 m

velocity = 19 m/s

angle =33.2 °

a) $v_x = v cos (\theta)$

   $v_x = 19 cos (33.2^0)$

   $v_x = 15.89\ m/s$

   $v_y = v sin (\theta)$

   $v_y = 19 sin (33.2^0)$

   $v_y = 10.40\ m/s$

magnitude

   $v = \sqrt{15.89^2+10.40^2}$

         v = 18.99 m/s

b) direction

         $tan \theta = \dfrac{10.40}{15.89}$

         $\theta = tan^{-1}(0.654)$

         $\theta = 33.20^0$