# Projectile fired down a slope

1. Apr 3, 2015

### Lord Anoobis

1. The problem statement, all variables and given/known data
An archer standing on a 15 degree slope shoots an arrow 20 degrees above the horizontal. How far down the slope does the arrow hit if it is shot with a speed of 50m/s from 1.75m above the ground?

2. Relevant equations

$y = x$tan$\theta$ - $\frac{gx^2}{2v^2 cos^2\theta}$

3. The attempt at a solution
I took the slope the archer is standing on to be a line through the origin of the x-y coordinate system, thus intersecting the trajectory of the arrow at two points, one being (0, 0). The slope this line is then

$y = -(tan15^o)x$

Setting the equations equal to each other and punching in values I obtained

$-(tan15^o)x = xtan20^o - \frac{9.8x^2}{5000cos^2 20^o}$

$x = \frac{(tan15^o + tan20^o)(5000cos^2 20^o)}{9.8}$

with x = 284.7 and y = -76.28 giving a final distance of 294.7m. The actual answer is 297m according to the book. I've been over this calculation a few times and it is clear that another set of eyes is required to shed some light on this conundrum. Please assist.

2. Apr 3, 2015

### TSny

Did you take care of the fact that the arrow is shot from 1.75 m above the ground?

3. Apr 3, 2015

### rhino1000

A=g.
initial upward velocity = sin(20)*50.
initial y = 1.75
x=cos(20)*50*t
y=-.5*a*t^2+vo*t+yo
yarrow=-.5*9.8*t^2+sin(20)*50*t+1.75

yslope= mx+b
yslope= tan(-15)*x+0
yslope=tan(-15)*cos(20)*50*t

yslope=yarrow
tan(-15 deg)*cos(20 deg)*50*t=-.5*9.8*t^2+sin(20 deg)*50*t+1.75

t=6.118s.
x=cos(20)*50*6.118
x=287

I tried. It's possible the book has a typo.

4. Apr 3, 2015

### Staff: Mentor

You don't seem to have taken into account the initial height of 1.75 m.

(TSny beat me to it.)

5. Apr 3, 2015

### Lord Anoobis

It seems I did not, and shifting the curve upward by 1.75m does indeed solve the problem.
Something else though. Initially I thought about rotating the system so that the ground is level and the launch angle is 35 degrees. Am I correct that this would not work? If one considers say, the same 15 degree slope and a launch angle of 75 degrees, rotating would simply send the arrow straight up, not so?

6. Apr 3, 2015

### TSny

Right. It would not work unless you also rotated the direction of the acceleration of gravity.

That's right (assuming gravity still acts vertically)

7. Apr 3, 2015

### Lord Anoobis

Of course. I didn't consider that, but I can see now that going about it that way wouldn't be simpler at all, seeing that compensating for the direction of gravitational acceleration would result in a horizontal component thereof as well.
Thanks for the help.