Arrow shot at 20 degrees on a 15 degree slope

[Lord Anoobis]

Homework Statement

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?

Homework Equations

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

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.

 

[TSny]

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

 

[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.

 

[Doc Al]

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.

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

(TSny beat me to it.)

 

[Lord Anoobis]

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

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?

 

[TSny]

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?

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

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?

That's right (assuming gravity still acts vertically)

 

[Lord Anoobis]

Right. It would not work unless you also rotated the direction of the acceleration of gravity.
That's right (assuming gravity still acts vertically)

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.