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Sagot :
Answer:
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°
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