Question

After leaving the end of a ski ramp, a ski jumper lands downhill at a point that is displaced 68.3 m horizontally from the end of the ramp. His velocity, just before landing, is 29.3 m/s and points in a direction 33.6 ° 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.

Answer 1

Answer:

a)52.58 m/s

b)56.13°

Explanation:

assume the upward direction as positive

x-component of the velocity = 29.3×cos33.6°=24.40 m/s (remain constant)

y-component of the velocity which is -29.3sin33.6°= -16.21 m/s

time of flight = 68.3/24.40= 2.7991 seconds

now, we can obtain final velocity in y-direction

$v_f_y= v_i_y-gt$

$v_f_y=-16.21-(-9.8)×2.7991$

=43.66 m/s

$v_0=\sqrt{29.3^2+43.66^2}$

=52.58 m/s

for direction

$tan^{-1}\frac{43.66}{29.3}$

56.13° from the horizontal