Question

After leaving the end of a ski ramp, a ski jumper lands downhill at a point that is displaced 50.0 m horizontally from the end of the ramp. His velocity, just before landing, is 24.0 m/s and points in a direction 34.1 ° 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

Consider the motion along the horizontal direction

x = horizontal displacement = 50 m

vox = initial velocity = ?

vfx = final velocity = 24 Cos34.1

ax = acceleration = 0 m/s2 (Since there is no force along the x-direction)

Using the kinematics equation

vfx2 = vox2 + 2 ax x

(24 Cos34.1)2 = vox2 + 2 (0) (50)

vox = 24 Cos34.1

Consider the motion along the vertical direction :

y = vertical displacement = 50 m

voy = initial velocity = ?

vfy = final velocity = - 24 Sin34.1

ay = acceleration = - 9.8 m/s2

Using the kinematics equation

vfy2 = voy2 + 2 ay y

(- 24 Sin34.1)2 = voy2 + 2 (- 9.8) (0)

voy = 24 Sin34.1

magnitude of the initial velocity is given using Pythagorean theorem as

vo = sqrt(vox2 + voy2) = sqrt((24 Cos34.1)2 + (24 Sin34.1)2) = 24 m/s

b)

= Direction = tan-1(voy/vox) = tan-1(24 Sin34.1/(24 Cos34.1))

= 34.1 deg