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## 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.