How Does Projectile Motion Apply in Olympic Ski Jumping?

[nerdmon]

Homework Statement

In the 2006 Winter Olympics, Thomas Morgenstern of Austria won a gold medal in one of the ski jumping events; his longest jump was R=140.0 m.

He takes off from point A at the top of a mountain at an angle of 15.0 degrees above the horizontal, with an initial speed of v0 = 26.9 m/s. His trajectory is that of a parabolic pathway, in which he lands at point B. The landing hill is curved and is quite steep near point B. You should ignore air resistance in this problem.

a) How long is he in the air? (From point A to B)
b) Find the magnitude and the direction of his velocity just before he lands at B
c) Based on your answer to part (b), explain why it is safer for ski jumpers to land on a steep slope than on a flat surface. (Hint: a large sudden change in velocity requires a large impact force.)

Homework Equations

deltax=v0t + 1/2at^2
uhh i don't know what else to use

The Attempt at a Solution

for a i did:
140=26.9t
t = 5.20 sec (is this right?)
and then i didn't know how to do b

 

[andrevdh]

For a) you need to use the horizontal speed, not the initial speed, which is directed fifteen degrees above the horizontal.

 

[nlightner]

or c, sorry

Hi, thank you for your question. I would like to provide you with a detailed and accurate response to help you understand and solve this problem. Firstly, let's start with the basic definition of projectile motion.

Projectile motion is a form of motion in which an object or particle (in this case, the ski jumper) is thrown near the Earth's surface, and it moves along a curved path under the action of gravity. This type of motion is typically studied in physics and is important in various sports, including ski jumping.

Now, let's address the questions in the homework statement.

a) To determine the time the ski jumper is in the air, we can use the equation d = v0t + 1/2at^2, where d is the distance travelled, v0 is the initial velocity, a is the acceleration (in this case, due to gravity), and t is the time. We know that the distance travelled is 140m, the initial velocity is 26.9m/s, and the angle of projection is 15 degrees. Since we are ignoring air resistance, the only acceleration acting on the jumper is due to gravity, which is 9.8m/s^2. Thus, the equation becomes:

140 = 26.9t cos 15 + 1/2(9.8)t^2
Simplifying this equation, we get a quadratic equation: 4.9t^2 - 26.9t cos 15 + 140 = 0
Solving for t using the quadratic formula, we get t = 5.20s or t = 5.72s. Since we are only interested in the time the jumper is in the air, we can disregard the second solution, which is a negative value. Therefore, the answer to part a is t = 5.20s.

b) To find the magnitude and direction of the jumper's velocity just before landing at B, we can use the equations for velocity in the x and y direction: vx = v0 cos θ and vy = v0 sin θ - gt, where θ is the angle of projection and g is the acceleration due to gravity. We know that the initial velocity is 26.9m/s and the angle of projection is 15 degrees. Thus, the magnitude of the velocity just before landing is:

v = √(vx