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1. The problem statement, all variables and given/known data

This is one of the last parts of a bigger problem, but I get the first parts. So:

The projectile motion describes a skier's jump. The initial velocity is 21.469m/s at 20 degrees to the horizontal. Find the maximum height the skier reaches (relative to the take off point), find distance 'd' down the slope where the skier lands.

3. The attempt at a solution

First I wrote the equations:

Vx = 21.469cos(20)

Vy= 21.469sin(20)-gt

Then, I found the time it took for the skier to reach the top:

Vy=21.469sin(20)-9.81t

9.81t=21.469sin(20)

t= 0.7485 seconds

Then I found the 'h':

ds/dt = 21.469sin(20)-9.81t

*integrating...*

s=21.469sin(20)t-4.905t^{2}

s(0.7485) = 2.7481 m

Then I drew this:

On it, I calculated the 30.2 value by plugging in 0.7485*2 into an x-direction distance equation:

Sx = 0.7485*2 x 21.469cos20 = 30.2

I then found the 17.44 using just trig.

So now, I started to treat the big triangle (described by p and q) as a similar triangle with sides 30.2&17.44 and developed a relationship:

q/p = 17.44/30.2

q=0.577p

Next, I expressed d in terms of p and q:

d^{2}=p^{2}+q^{2}

d^{2}= 1.3329p^{2}

and thus, p=0.866d

I performed similar operations and got:

q=0.5d

Then time for the skier to go the whole horizontal distance:

0.886d = 21.469cos(20)*t

t=0.04293d

And thus the time it will take for the skier to hit the ground from her highest point is:

t=0.0429d-0.7485

Now, we can plug in that equation into the equation of vertical distance travelled:

h+0.5d = 21.469sin(20)(0.0429d-0.7485)-4.905(0.04293d-0.7485)^{2}

I can simplify that and try to find d (from discriminant and things) but I end up with a negative discriminant... Help?

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# Homework Help: Projectile motion problem

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