Physics, 25.09.2019 20:20 leethiel1

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’?

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Answer from: ruby562

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:

$\mathbf{= 110 \ \times (\dfrac{1000 \ m}{3600 \ s})}$

= 30.556 m/s

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

$\mathbf{V_{horizontal} = 30.556 \ m/s}$ $\mathbf{V_{vertical } = 0 \ m/s}$

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:

$\mathbf{S_v= ut + \dfrac{1}{2} gt^2}$

where the initial velocity = 0 m/s

$\mathbf{S_v= (0)t + \dfrac{1}{2} gt^2}$

$\mathbf{S_v= \dfrac{1}{2} gt^2}$

$\mathbf{S_v= \dfrac{1}{2} (9.81)t^2}$

$\mathbf{S_v=4.905t^2}$

The horizontal distance $\mathbf{S_x = ut }$

$\mathbf{S_x = 30.556t }$

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

$\mathbf{tan \ \theta = \dfrac{opposite}{adjacent}}$

where;

θ = 45°

$\mathbf{tan \ 45^0 = \dfrac{S_v}{S_x}}$

$\mathbf{tan \ 45^0 = \dfrac{4.905t^2}{30.556t}}$

1 × 30.556t = 4.905t²

Divide both sides by t, we have:

30.556 = 4.905t

$\mathbf{t = \dfrac{30.556}{4.905}}$

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.

$\mathbf{S_v=4.905(6.229)^2}$ $\mathbf{S_v=4.905\times 38.800}$ $\mathbf{S_v=190.314 \ m}$

The horizontal distance $\mathbf{S_x = 30.556t }$

$\mathbf{S_x = 30.556(6.229) }$ $\mathbf{S_x = 190.33\ m}$

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

= $\mathbf{190.314 \sqrt{2}\ m}$

= 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:

link

In a flying ski jump, the skier acquires a speed of 110 km/h by racing down a steep hill and then li

Answer from: samariamartinez1029

Approximately $\displaystyle\rm \left[ \begin{array}{c}\rm191\; m\\\rm-191\; m\end{array}\right]$ .

Explanation:

Consider this $45^{\circ}$ 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 (0, 0) .

Assume that the skier is running in the positive x-direction. The line that represents the slope shall point downwards at $45^{\circ}$ 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 y = -x .

Convert the initial speed of this diver to SI units:

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

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 ( $g \approx \rm -9.81\; m\cdot s^{-2}$ near the surface of the earth.) At seconds after the skier goes beyond the edge of the ramp, the position of the skier will be:

-coordinate: 30.556t

meters (constant velocity;)

-coordinate: $\displaystyle -\frac{1}{2}g\cdot t^{2} = -\frac{9.81}{2}\cdot t^{2}$ meters (constant acceleration with an initial vertical velocity of zero.)

To eliminate from this expression, solve the equation between and for . That is: express as a function of .

$x = 30.556\;t\implies \displaystyle t = \frac{x}{30.556}$ .

Replace the in the equation of with this expression:

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

Plot the two functions:

, $\displaystyle y= -0.0052535\;x^{2}$ ,

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

Alternatively, equate the two expressions of (right-hand side of the equation, the part where is expressed as a function of .)

$-0.0052535\;x^{2} = -x$ ,

$\implies x = 190.35$ .

The value of can be found by evaluating either equation at this particular -value: x = 190.35 .

y = -190.35 .

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