In a flying ski jump, the skier acquires a speed of 110 km/h by racing down a steep hill and then lifts off into the air from a horizontal ramp. Beyond this ramp, the ground slopes downward at an angle of 45◦. (a) Assuming that the skier is in a free-fall motion after he leaves the ramp, at what distance down the slope will he land? What is his displacement vector from the point of ‘lift off’?

Consider this [tex]45^{\circ}[/tex] slope and the trajectory of the skier in a cartesian plane. Since the problem is asking for the displacement vector relative to the point of "lift off", let that particular point be the origin [tex](0, 0)[/tex].

Assume that the skier is running in the positive x-direction. The line that represents the slope shall point downwards at [tex]45^{\circ}[/tex] to the x-axis. Since this slope is connected to the ramp, it should also go through the origin. Based on these conditions, this line should be represented as [tex]y = -x[/tex].

Convert the initial speed of this diver to SI units:

[tex]\displaystyle v = \rm 110\; km\cdot h^{-1} = 110 \times \frac{1}{3.6} = 30.556\; m\cdot s^{-1}[/tex].

The question assumes that the skier is in a free-fall motion. In other words, the skier travels with a constant horizontal velocity and accelerates downwards at [tex]g[/tex] ([tex]g \approx \rm -9.81\; m\cdot s^{-2}[/tex] near the surface of the earth.) At [tex]t[/tex] seconds after the skier goes beyond the edge of the ramp, the position of the skier will be:

[tex]x[/tex]-coordinate: [tex]30.556t[/tex] meters (constant velocity;)
[tex]y[/tex]-coordinate: [tex]\displaystyle -\frac{1}{2}g\cdot t^{2} = -\frac{9.81}{2}\cdot t^{2}[/tex] meters (constant acceleration with an initial vertical velocity of zero.)

To eliminate [tex]t[/tex] from this expression, solve the equation between [tex]t[/tex] and [tex]x[/tex] for [tex]t[/tex]. That is: express [tex]t[/tex] as a function of [tex]x[/tex].

[tex]x = 30.556\;t\implies \displaystyle t = \frac{x}{30.556}[/tex].

Replace the [tex]t[/tex] in the equation of [tex]y[/tex] with this expression:

[tex]\begin{aligned} y = &-\frac{9.81}{2}\cdot t^{2}\\ &= -\frac{9.81}{2} \cdot \left(\frac{x}{30.556}\right)^{2}\\&= -0.0052535\;x^{2}\end{aligned}[/tex].

Plot the two functions:

[tex]y = -x[/tex],
[tex]\displaystyle y= -0.0052535\;x^{2}[/tex],

and look for their intersection. Refer to the diagram attached.

Alternatively, equate the two expressions of [tex]y[/tex] (right-hand side of the equation, the part where [tex]y[/tex] is expressed as a function of [tex]x[/tex].)

[tex]-0.0052535\;x^{2} = -x[/tex],

[tex]\implies x = 190.35[/tex].

The value of [tex]y[/tex] can be found by evaluating either equation at this particular [tex]x[/tex]-value: [tex]x = 190.35[/tex].

[tex]y = -190.35[/tex].

The position vector of a point [tex](x, y)[/tex] on a cartesian plane is [tex]\displaystyle \left[\begin{array}{l}x \\ y\end{array}\right][/tex]. The coordinates of this skier is approximately [tex](190.35, -190.35)[/tex]. The position vector of this skier will be [tex]\displaystyle\rm \left[ \begin{array}{c}\rm191\\\rm-191\end{array}\right][/tex]. Keep in mind that both numbers in this vectors are in meters.

ajeigbeibraheem ajeigbeibraheem · Answer 2 · 2021-10-20T10:21:28+02:00

In a flying ski jump, the skier acquires a speed of 110 km/h by racing down a steep hill and then lifts off into the air from a horizontal ramp. Assuming that the skier is in a free-fall motion, the distance down the slope that he will land is 190.3 m and the total displacement vector from the point of lift-off is 269.15 m

From the given information;

the speed of the skier V = 110 km/hr

To convert the speed into m/s, we have:

[tex]\mathbf{= 110 \ \times (\dfrac{1000 \ m}{3600 \ s})}[/tex]

= 30.556 m/s

Since the speed race down the steep hill before flying horizontally into the air, then we can assert that the;

[tex]\mathbf{V_{horizontal} = 30.556 \ m/s}[/tex]
[tex]\mathbf{V_{vertical } = 0 \ m/s}[/tex]

Now, for a motion under free-fall, to determine the distance(S) of the slope using the second equation of motion, we have:

The distance along the vertical axis to be:

[tex]\mathbf{S_v= ut + \dfrac{1}{2} gt^2}[/tex]

where the initial velocity = 0 m/s

[tex]\mathbf{S_v= (0)t + \dfrac{1}{2} gt^2}[/tex]

[tex]\mathbf{S_v= \dfrac{1}{2} gt^2}[/tex]

[tex]\mathbf{S_v= \dfrac{1}{2} (9.81)t^2}[/tex]

[tex]\mathbf{S_v=4.905t^2}[/tex]

The horizontal distance [tex]\mathbf{S_x = ut }[/tex]

[tex]\mathbf{S_x = 30.556t }[/tex]

From the angle at which the ground begins to slope, taking the tangent of the angle with respect to the distance, we have:

[tex]\mathbf{tan \ \theta = \dfrac{opposite}{adjacent}}[/tex]

where;

θ = 45°

[tex]\mathbf{tan \ 45^0 = \dfrac{S_v}{S_x}}[/tex]

[tex]\mathbf{tan \ 45^0 = \dfrac{4.905t^2}{30.556t}}[/tex]

1 × 30.556t = 4.905t²

Divide both sides by t, we have:

30.556 = 4.905t

[tex]\mathbf{t = \dfrac{30.556}{4.905}}[/tex]

t = 6.229 sec

However, since the time at which the slope will land is known, we can easily determine the value of the vertical and horizontal distances.

i.e.

[tex]\mathbf{S_v=4.905(6.229)^2}[/tex]
[tex]\mathbf{S_v=4.905\times 38.800}[/tex]
[tex]\mathbf{S_v=190.314 \ m}[/tex]

The horizontal distance [tex]\mathbf{S_x = 30.556t }[/tex]

[tex]\mathbf{S_x = 30.556(6.229) }[/tex]
[tex]\mathbf{S_x = 190.33\ m}[/tex]

The total displacement vector from the point of lift-off is:

= [tex]\mathbf{190.314 \sqrt{2}\ m}[/tex]

= 269.15 m

Therefore, we can conclude that the distance down the slope that he will land is 190.3 m and the displacement down the slope that the skier will land is 269.15 m

Learn more about displacement here:

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