Secret agent Samantha Bond, 37.00 kg, has Just ignited her jet powered skis at t

Question

Secret agent Samantha Bond, 37.00 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 210.0 N. when 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

Figure out Bond's vertical and horizontal accelerations when she leaves the ground.

At F = 210.0 N of thrust at an angle of 10 degree from horizontal, Bond's skis provide her with a horizontal force of

Fh = F cos ((210.0 N) (cos 10 degree) 206.8 N)

and a vertical force of

Fv = F sin (= (210.0 N) (sin 10 degree) 36.46 N upward)

Bond weighs mg (= (36.00 kg) (9.81 N/kg) = 353.16 N downward),

so her net vertical force after she takes the jump is

F(net vertical) = mg - Fv

and therefore her net downward vertical acceleration av is

av = F(net downward) / m

av = (mg - Fv) / m

av = g - Fv / m

Her net horizontal acceleration ah is just

ah = Fh / m

2. Find out how fast she is traveling at that moment.

At the foot of the slope, Bond is acted on by two forces parallel with the slope:

Gravity pulls her down the slope with a force of F(backward) = mg sin .

Her jet skis push her up the slope with a force of F(jetski) = 210.0 N.

So her net force up the slope is F(jetski) - mg sin , making her acceleration equal to

a(slope) = F(slope) / m

= [F(jetski) - mg sin ] / m

= F(jetski) / m - g sin .

She travels at this acceleration for the distance of the slope, which is easily found by sketching the slope and using trigonometry:

sin = (50 m) / x (the slope length), so slope length x = (50 m) / sin

To find the velocity, use the equation v^2 = v0^2 + 2ax and realize that v0 is zero (we assume Bond starts from rest). So v^2 = 2ax, and therefore

v(slope) = sqrt [2a(slope)x]
= sqrt [2 ( F(jetski) / m - g sin ) (50 m) / sin ]
= sqrt [ (100 m) ( F(jetski) / (m sin ) - g ) ]

3. Use the above information to figure out long Bond spends in the air.

Here is what we know so far:
Fh = F cos
Fv = F sin
av = g - Fv / m
ah = Fh / m
slope length x = (50 m) / sin

v(slope) = sqrt [ (100 m) ( F(jetski) / (m sin ) - g ) ]

Splitting this initial velocity into horizontal (vh) and vertical (vv) components yields:

vh = v(slope) cos
= sqrt [ (100 m) ( F(jetski) / (m tan ) - g cos ) ]

vv = v(slope) sin
= sqrt [ (100 m) ( F(jetski) / m - g sin ) ] in the UPWARD direction

The general equation of motion is

0 = x + (v0) t + 0.5 * at^2

Solving for t (use the quadratic equation):

t = ( -v0 +- sqrt[v0^2 - 4(0.5 * a)(x)] ) / 2(0.5 * a)
= ( -v0 +- sqrt[v0^2 - 2ax] ) / a

where v0 is the initial speed, vv, in the UPWARD direction and a is av in the DOWNWARD direction. In other words, if we are considering the DOWNWARD drop and acceleration to be positive, then v0 = -vv, and

t = ( vv +- sqrt[vv^2 - 2(av)x] ) / av

Ignore the - part of +-, since the represents the mathematical "landing" from before Bond took off.

t = ( vv + sqrt[vv^2 - 2(av)x] ) / av
= ( sqrt [ (100 m) ( F(jetski) / m - g sin ) ] + sqrt [ [ (100 m) ( F(jetski) / m - g sin ) ] - 2(av)x] ) / av

(if you're getting tired of carrying around all these equations, you can find a numerical value for t by plugging into the above equation, where you find that

t = ( sqrt [ (100 m) ( F(jetski) / m - g sin ) ] + sqrt [ [ (100 m) ( F(jetski) / m - g sin ) ] - 2(g - Fv / m)x] ) / (g - Fv / m)

t = ( sqrt [ (100 m) ( (210.0 N) / (36.00 kg) - (9.81 N/m) sin 10 degree ) ] + sqrt [ [ (100 m) ( (210.0 N) / (36.00 kg) - (9.81 N/m) sin 10 degree ) ] - 2((9.81 N/m) - ((210.0 N) sin 10 degree) / (36.00 kg)) [ (50 m) / sin 10 degree ] ] ) / ((9.81 N/m) - ((210.0 N) sin 10 degree) / (36.00 kg))

t = 26.61 s

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