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

draw a picture first

Like the triangle?

|dw:1380722857043:dw|

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

leme draw wait

sint it like that?!

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

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

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

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

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

how did you get that number!?

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

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

So wait how do I solve for each velocity?

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

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, \[v_0=25.81\ m/s\] \[x\approx 34\ m\]

Hmmm what was the answer before you rounded?

33.99 m for the x

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

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.

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

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.

Is it ok?

Hold on let me check

\[x=2v^2_0\sinθ\cosθ/g≈68 m\]

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

Why do we use this formula instead?

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.

Now, it should be OK.

ok thanks :)

;)

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). ;)

ha I'll let you know :)

It was correct thank you :)

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