Question

Secret agent Samantha Bond, 48.50 kg, has just ignited her jet powered skis at the foot of a 10.0 degree slope that rises in height by 50.0 m. The skis have a thrust of 244.0 N. When she gets to the top of the slope, she keeps the skis at 10.0 degrees as she flies through the air. At what horizontal distance from the top of the slope, from where she took off, will she land? This will be 50.0 m below where she took off. The thrust of the skis is parallel to the skis. There is no friction on the slope, and air resistance can be ignored.

Answer 1

Horizontal Component of Force Fh = F cos θ (= (244.0 N) (cos 10°) ≈ 240.3 N
Vertical Component of Force Fv = F sin θ (= (244.0 N) (sin 10°) ≈ 42.4 N upward
F(net vertical) = mg - Fv = 48.5 *10 - 42.4 = 442.6 N


Vertical acceleration av = Fnet / m (downward direction) = 442.6 / 48.5 = 9.12 m/s^2
Horizontal acceleration ah = Fh / m = 240.3 / 48.5 = 4.95 m/s^2

During Slope a = F - mg sin(10) / m
a slope = (244 - 485 * sin(10)) / 48.5 = 3.3 m/s^2

Initial Velocity = 0
Final Velocity is given by v^2 = u^2 + 2*a*s
where s = 50/sin(10)
s = 288 m
v^2 = 2 *3.3 * 288
v = 43.6 m/s

Horizontal component vh = v * cos(10) = 43.6 * cos(10) = 42.9 m/s
Vertical component vv = v * sin(10) = 43.6 * sin(10) = 7.6 m/s

v = u - av*t
t1 = 7.6 / 9.12 s = 0.83s

v^2 = u^2 -2*av*h
h = 7.6^2 / 2 *9.12
h =  3.16 m
Now she starts falling down, Total height = 50 + 3.16
53.16 = 0.5 * 9.12 * t^2
t = 3.41 s

Total time in the air = 3.41 +0.83 s = 4.24s

S = 42.9*4.24 + 0.5 * 4.95 * 4.24^2
S = 226.4 m
horizontal distance from the top of the slope, from where she took off, will she land S = 226.4 m