A ski jumper travels down a slope and leaves the ski track moving in the horizontal direction with a speed of 28 m/s as in the figure.

The landing incline below her falls off with a

slope of θ = 45◦

.The acceleration of gravity is 9.8 m/s^2

Calculate the distance d she travels along

the incline before landing.

Answer in units of m.
Determine how long the ski jumper is airborne.
Answer in units of s.
What is the magnitude of the relative angle φ
with which the ski jumper hits the slope?
Answer in units of ◦

Explanation:

Nothing happens unless we know the position relationship between launch ramp and landing slope. That missing figure would sure be helpful.

ASSUMING the landing slope starts its 45° slope at the position where the skier leaves the launch track. Also assuming no air resistance or lift due to airfoil action on body and skis. (projectile trajectory)

With the launch point as origin and original velocity in the positive direction, horizontal position can be described as

x = 28t

With the launch point as origin and DOWN being positive direction, vertical position can be described as

y = ½(9.8)t²

y = 4.9t²

On a 45° downward slope, y will increase at the same rate as x. time t is common to both equations. When our y equals our x, landing occurs.

4.9t² = 28t

4.9t = 28

t = 5.714285...s

t = 5.71 s

The horizontal and vertical distances at landing are

x = 28(5.714285) = 160 m

so along the slope, the distance is

d = √(2(160²)) = 226.2741699...

d = 226 m

velocities will give us the angle of flight, subtract the slope.

φ = arctan(gt/vx) - 45 = arctan(9.8(5.714285) / 28) - 45 = 18.434948...

φ = 18°