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.

Answer from: Quest

Is a vector quantity that refers to "the rate at which an object changes its position." imagine a person moving rapidly - one step forward and one step back - always returning to the original starting position. while this might result in a frenzy of activity, it would result in a zero velocity. because the person always returns to the original position, the motion would never result in a change in position. since velocity is defined as the rate at which the position changes, this motion results in zero velocity. if a person in motion wishes to maximize their velocity, then that person must make every effort to maximize the amount that they are displaced from their original position. every step must go into moving that person further from where he or she started. for certain, the person should never change directions and begin to return to the starting position.

Answer from: Quest

beta is electron from nucleus

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