- #1

Melawrghk

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**Projectile motion problem - SOLVED**

## Homework Statement

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.

## 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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