Question

A ski jumper starts from rest 44.8 m above
the ground on a frictionless track and flies
off the track at an angle of 44.4 above the
horizontal and at a height of 10.2 m above the
level ground. Neglect air resistance.
The acceleration of gravity is 9.8 m/s2 .
What is his speed when he leaves the track?
Answer in units of m/s

max altitude is 27.1377

Where does he land relative to the end of the
track? answer in units m

Answer 1

draw a picture first

Answer 2

Like the triangle?

Answer 3

no the entire diagram.. the track.. he starts 44.8m above ..

Answer 4

leme draw wait

Answer 5

sint it like that?!

Answer 6

Well that does make sense ha I wasn't provided a picture

Answer 7

oh.. gee... oki.. i spoilt it for you then :P.. sorry my bad.. now u can do right?

Answer 8

Ummm well all I need ro do is find how far he landed from the track but I haven't found an equation that gives me the correct answer

Answer 9

did you find the velocity with which he leaves the ramp?

Answer 10

Well the only velocity I have is 26.0415 but I'm not sure if it changes

Answer 11

how did you get that number!?

Answer 12

\[v=\sqrt{2g(h1-h2)}\]

Answer 13

ok.. ok.. thats right..

now its projectile motion.. 2d projectile motion.. split the velocities into horizontal and vertical components.. and solve

Answer 14

So wait how do I solve for each velocity?

Answer 15

by taking the components.. haven't you solved 2d projectile motion problems before? if no u better look taht up!

Answer 16

In order to find the velocity you can apply the theorem of conservation of energy,

\[U_1=K+U_2\]
\[mgh_1=mv_0^2/2+mgh_2\]
\[v_0=\sqrt{2g(h_1-h_2)}\]
where
\[h_1=44.8\ m\]\[h_2=10.8\ m\]\[g=9.8\ m/s^2\]
If you want to obtain the components of the velocity vector, then
\[v_{0x}=v_0\cos\theta\]\[v_{0y}=v_0\sin\theta\]
where
\[\theta=44.4\ \text{degrees}\]

This velocity will hep us in the last part. As you want to find the range of the jump, then, you can find the time where he reach its maximum altitude. In this point vy=0,
\[v_y=0=v_{0y}-gt\Rightarrow t=v_{0y}/g=v_0\sin\theta/g\]
Now that you find the time, you can find the range,
\[x=v_{0x}t=v_0\cos\theta t=v_0^2\sin\theta\cos\theta/g\]
Substitue the variables with the values given in the problem, and you'll find the solutions.

Only to check, I find that,

\[v_0=25.81\ m/s\]

\[x\approx 34\ m\]

Answer 17

Hmmm what was the answer before you rounded?

Answer 18

33.99 m for the x

Answer 19

Oh well was that using the velocity you found or mine?

Answer 20

It is not the total range, it's the range from the ramp to the point where the jumper reach again the level of the ramp. It's possible the problem ask you for the total range.

Answer 21

Better, I forget to put a factor 2 in the time I found.

Answer 22

Because without the 2 it is the time to reach the maximum altitude. With the 2 factor, it is the time to reach again the level of the ramp.

Answer 23

Is it ok?

Answer 24

Hold on let me check

Answer 25

\[x=2v^2_0\sinθ\cosθ/g≈68 m\]

Answer 26

Don't need to check. We need to find another distance.

Answer 27

Why do we use this formula instead?

Answer 28

Sorry!! My fault!! I put a cosine instead a sine.

You need to solve,
\[0=10.8+25.81\sin(44.4)t−\frac{1}{2}9.8t^2⇒t=4.21 s\]
Then
\[x=v_0\cos\theta t=25.81\cos(44.4)4.21=77.63 m\]

Because you need to land in the ground. If you want to find the range between the ramp and the point where the jumper will be at the same level of the ramp again, the we would use the formula I give you before.

But you need to land, so y=0, and you begin in 10.8 m.

Answer 29

Now, it should be OK.

Answer 30

ok thanks :)

Answer 31

;)

Answer 32

Don't forget to check the answer and, please, tell me if it is wrong (the formalism is ok but it is possible a I put some numbers off their place). ;)

Answer 33

ha I'll let you know :)

Answer 34

It was correct thank you :)