Secret agent Samantha Bond, 47.00 kg, has just ignited her jet powered skis at t

Question

Secret agent Samantha Bond, 47.00 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 234.0 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.

Explanation / Answer

We will calculate the launch speed at the top of the slope, then treat SB as a projectile with ongoing vertical and horizontal thrust from her skis.

Gravity force down the slope is m*g*sin(10) = 47*9.81*sin(10) = 80.063N

So accelerating force parallel to the slope = 234 – 80.063N = 153.936 N
So acceleration up the slope a = 153.936 N / 47 kg = 3.275 m/s^2
Distance up the slope s = 50m / sin(10) = 287.9m

So launch speed at the top of the slope is u = sqrt(2*a*s)
u = sqrt(2*3.275*287.9) = 43.42 m/s
Vertical component of u is uv = 43.42*sin(10) = 7.54 m/s
Horizontal component of u is uh = 43.42*cos(10) = 42.76 m/s

After launch, vertical upward thrust is 234*sin(10) N = 40.63N
giving upward acceleration of 40.63N / 47kg = 0.8645 m/s^2
So net downward acceleraton is g = 9.81 - 0.8645 = 8.945 m/s^2

Additional height reached after launch is h = uv^2 / 2g
h = 7.54^2 / 2*8.945 = 3.1778 m
taking time t(up) = uv/g = 7.54 / 8.945 = 0.843 s

SB now falls 50+3.1778m = 53.1778 m with downward accel 8.945 m/s^2
taking time t(down) = sqrt(2*53.1778 / 8.945) = 3.448 s

So SB's total flight time is t = 3.448 + 0.843 s = 4.2911 s

Horizontal thrust is 234*cos(10) N = 230.445 N
giving horizontal acceleration a = 230.445 N / 47kg = 4.903 m/s^2

So horizontal distance travelled is d = uh*t + 0.5*a*t^2

d = (42.76*4.2911) + 0.5*4.903*4.2911^2

d = 183.48 + 45.14 m = 228.62 m <= ANS

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