A skier leaves the rump of a ski jump with a velocity of v 16.0 m s = 29.00 abov

Question

A skier leaves the rump of a ski jump with a velocity of v 16.0 m s = 29.00 above the horizontal as shown in he figure. The slope where she will land is inclined downward at 50.0° and air resistance is negligible. (a) Find the distance from the end of the ramp to where the jumper lands. (b) Find the velocity components just berore the landing. (Let the positive x direction be to the right and the positive y direction be up.) m/s m/s (e) Explain how you think the results might be affected if air resistance were included? This answer has not been graded yet Need Help?adtTalk to a Tutor

Explanation / Answer

v_x = 16.0 m/s cos(29 deg) = 13.994 m/s
v_y0 = 16.0 m/s sin(29 deg) = 7.757 m/s
Max height = height when v_y = 0, can be found from
|2gh| = (v_y0)^2, so h_max = 3.07 meters
Time to reach max height = v_y0/g = 0.79 seconds

Seek a time when the skier will be 50.0 degrees below the starting point.
t = x / v_x
y = 3.07 - (1/2)(9.8 m/s^2) (t-0.79)^2
y/x = tan(-50 deg)
These are 3 equations in 3 unknowns.
Eliminate x from the 1st & 3rd equations:
y = (t v_x) tan(-50 deg) = -16.677 t
Substitute this result into the equation that was quadratic in t:
-16.677 t = 3.07 - 4.9 t^2 + 7.742 t- 3.06
4.9 t^2 - 24.42 t - 0.01= 0

Solving this quadratic equation, we get two values of t, positive and negative. We accept only positive as t cannot be negative. Therefore

t=4.98 seconds

x = t v_x = (4.98 s)(13.994 m) = 69.69 meters
"Slant" distance from ramp's end is
28.07 m divided by cos(50 deg) =108.42 meters

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.