An archer standing on a 15 degree slope shoots an arrow at an angle of 20 degrees above the horizontal. how far below its original point of release does the arrow hit if it is shot with a speed of 32 m/s from a height of 1.99 m above the ground?

The arrow will hit 101 m from the point of release.

Explanation:

The position vector of the arrow can be calculated as follows:

r = (x0 + v0 · t · cos α, y0 + v0 · t ·sin α + 1/2 · g · t²)

Where:

r = position vector at time "t"

x0 = initial horizontal position.

v0 = initial velocity.

t = time.

α = launching angle.

y0 = initial vertical position.

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).

Please, see the attached figure for a better understanding of the problem. In red is depicted the position vector of the arrow at the final time (r final). Notice that the origin of the frame of reference is located at the releasing point so that y0 and x0 = 0. The position vector at the final time will be:

r final = (rx final, -1.99 m)

Using the equation for "ry final", we can obtain the final time and then calculate "rx final" that is the horizontal distance traveled by the arrow. Notice that the launching angle relative to the level ground is the sum of the angle of the slope and the shooting angle (20° + 15° = 35°):

y = y0 + v0 · t ·sin α + 1/2 · g · t²    (y0 = 0 m)

-1.99 m = 32 m/s · t · sin 35° - 1/2 · 9.8 m/s² · t²

0 = -4.9 m/s² · t² + 32 m/s · t · sin 35° + 1.99 m

Solving the quadratic equation:

t = 3.85 s

Now, with this time we can calculate how far the arrow hit:

x = x0 + v0 · t · cos α     (x0 = 0)

x = 32 m/s · 3.85 s · cos 35°

x = 101 m

The arrow will hit 101 m from the point of release.

area of focus