Question 4.5: The End of the Ski Jump A ski jumper leaves the ski track mo...
The End of the Ski Jump
A ski jumper leaves the ski track moving in the horizontal direction with a speed of 25.0 m/s as shown in Figure 4.15. The landing incline below her falls off with a slope of 35.0°. Where does she land on the incline?
Conceptualize We can conceptualize this problem based on memories of observing winter ski jumping competitions. We estimate the skier to be airborne for perhaps 4 s and to travel a distance of about 100 m horizontally. We should expect the value of d, the distance traveled along the incline, to be of the same order of magnitude.
Categorize We categorize the problem as one of a particle in projectile motion. As with other projectile motion problems, we use the particle under constant velocity model for the horizontal motion and the particle under constant acceleration model for the vertical motion.
Analyze It is convenient to select the beginning of the jump as the origin. The initial velocity components are v_{xi} = 25.0 m/s and v_{yi} = 0. From the right triangle in Figure 4.15, we see that the jumper’s x and y coordinates at the landing point are given by x_f = d cos Φ and y_f = -d sin Φ.
Express the coordinates of the jumper as a function of time, using the particle under constant velocity model for x and the position equation from the particle under constant acceleration model for y:
(1) x_f=v_{x i} t \rightarrow \quad (2) \quad d \cos \phi=v_{x i} t
(3) y_f=v_{y i} t-\frac{1}{2} g t^2 \rightarrow \quad (4) \quad -d \sin \phi=-\frac{1}{2} g t^2
Solve Equation (2) for t and substitute the result into Equation (4):
-d \sin \phi=-\frac{1}{2} g\left(\frac{d \cos \phi}{v_{x i}}\right)^2Solve for d and substitute numerical values:
d=\frac{2 v_{x i}{}^2 \sin \phi}{g \cos ^2 \phi}=\frac{2(25.0 m/s)^2 \sin 35.0^{\circ}}{\left(9.80 m/s^2\right) \cos ^2 35.0^{\circ}}=109 mEvaluate the x and y coordinates of the point at which the skier lands:
\begin{aligned}& x_f=d \cos \phi=(109 m) \cos 35.0^{\circ}=89.3 m \\& y_f=-d \sin \phi=-(109 m) \sin 35.0^{\circ}=-62.5 m\end{aligned}Finalize Let us compare these results with our expectations. We expected the horizontal distance to be on the order of 100 m, and our result of 89.3 m is indeed on this order of magnitude. It might be useful to calculate the time interval that the jumper is in the air and compare it with our estimate of about 4 s.