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.

air exerts forces on falling objects near earth’s surface.

explanation:

the term 'free falls' refer to object that are falling towards the earth, in a situation in which gravity is the only force acting on the object.

for an object in free fall, the acceleration is constant (g, the gravitational acceleration), and so the velocity of the object constantly increases.

for objects near earth's surface, there is always another force acting on the object: the resistance due to the air. this force is opposed to the direction of fall of the object, and it is proportional to the velocity of the object, so as the object falls down, this force increases up to a point when it balances the force of gravity, and the object continues its motion at constant velocity (called terminal velocity). since in this case gravity is not the only force acting on the object, we are not in a situation of free fall.