Question

A skier leaves the ramp of a ski jump with a velocity of $\mathit{v}\mathbf{=}\mathbf{10}\mathbf{.}\mathbf{0}\mathbf{m}\mathbf{/}\mathbf{s}$at $\mathit{\theta }\mathbf{=}\mathbf{15}\mathbf{.}{\mathbf{0}}^{\mathbf{\circ }}$ above the horizontal as shown in Figure . The slope where she will land is inclined downward at $\mathit{\varphi }\mathbf{=}{\mathbf{50}}^{\mathbf{\circ }}$, and air resistance is negligible. Find (a) the distance from the end of the ramp to where the jumper lands and (b) her velocity components just before the landing. (c) Explain how you think the results might be affected if air resistance were included.

Short answer

a) The distance from the end of the ramp to where the jumper lands are $44.6\mathrm{m}/\mathrm{s}$.

b) Her velocity components just before landing is ${\mathrm{v}}_{\mathrm{yf}}=-26.3\mathrm{m}/\mathrm{s}$, ${\mathrm{v}}_{\mathrm{xf}}=9.66\mathrm{m}/\mathrm{s}$.

c) She'll be carried far enough by the air resistance to extend her terrestrial range.

Step-by-step solution

 Step 1: Definition of velocity.

The rate of change of an object's location with regard to a frame of reference is its velocity, which is a function of time. A statement of an object's speed and direction of motion is referred to as velocity.

Find the distance from the end of the ramp to where the jumper lands.

(a)

When the skier exits the ramp, she falls like a projectile with a starting velocity of $v_i=10\text{m/s}$and an angle of $\theta ={15}^{\circ }$, thus the components of the are as follows:

$\begin{array}{c}{v}_{xi}=v_i\cos \theta \\ =10.0\cos 15°\\ =9.66\mathrm{m}/\mathrm{s}\\ v_{yi}=v_i\sin \theta \\ =10.0\sin 15°\\ =2.59\mathrm{m}/\mathrm{s}\end{array}$

And, because she falls on an incline with a downward slope of $\varphi =50.0°$, by taking the original at the end of the ramp,

$\begin{array}{c}{x}_i=0\\ x_f=d\cos \varphi \\ x_f=d\cos 50°\\ y_i=0\\ y_f=-d\sin \varphi \\ y_f=-d\sin 50°\end{array}$

Where $d$is the distance from the ramp to where she falls on the incline.

To calculate the time, she takes to land, we insert the supplied values into the kinematic equation as follows:

$\begin{array}{c}{x}_f=x_i+v_{x}t+0.5a_x{t}^{2}d\cos 50°\\ d=0+(9.66\mathrm{m}/\mathrm{s})t+0t\\ t=\frac{d\cos 50°}{9.66\mathrm{m}/\mathrm{s}}\\ t=0.066d{\left(\mathrm{m}/\mathrm{s}\right)}^{-1}\end{array}$

The following values are substituted into the kinematic equation for the vertical motion:

$\begin{array}{c}{y}_f=y_i+v_{yi}t+0.5a_y{t}^2-d\sin 50°\\ 0=0+(2.59\mathrm{m}/\mathrm{s})t+0.5(-9.8\mathrm{m}/{\mathrm{s}}^2)t^2\\ 0.766d=\left(2.59\mathrm{m}/\mathrm{s}\right)t-\left(4.9\mathrm{m}/{\mathrm{s}}^2\right)t^2\end{array}$

Substitute the above-mentioned $t$ value,

$\begin{array}{c}0.766d=2.59\mathrm{m}/\mathrm{s}\times (0.066d{\left(\mathrm{m}/\mathrm{s}\right)}^{-1})-4.9\mathrm{m}/{\mathrm{s}}^2(0.066d{\left(\mathrm{m}/\mathrm{s}\right)}^{-1}{)}^2{d}^2\\ 0.766d=0.171m-0.021d\\ d=44.6\mathrm{m}\end{array}$

Find her velocity components just before the landing.

(b)

Now figure out how long it will take her to land:

$\begin{array}{c}t=0.066d{\left(\mathrm{m}/\mathrm{s}\right)}^{-1}\\ =0.066(44.6s)\times {\left(\mathrm{m}/\mathrm{s}\right)}^{-1}\\ =2.94\mathrm{s}\end{array}$

Substitute the values into the kinematic equation for the vertical component of the

final velocity.

$\begin{array}{c}{v}_{yf}=v_{yi}+a_{y}t\\ =2.59\mathrm{m}/\mathrm{s}+(-9.8\mathrm{m}/{\mathrm{s}}^2)·(2.94s)\\ =-26.3\mathrm{m}/\mathrm{s}\end{array}$

Since horizontal motion has a constant velocity, the horizontal component of final velocity is as follows:

${\mathrm{v}}_{\mathrm{xf}}=9.66\mathrm{m}/\mathrm{s}$

Step 4: Explanation of how you think the results might be affected if air resistance were included.

(c)

The air resistance will limit her velocity in the motion direction, thus as she climbs, gravitation and air resistance will be working against her, lowering her maximum height. And because she lands below the spot where she started, the air resistance will carry her far enough to extend her landing range.