At the other end of what would come to be known as the Roman Empire, Parthian (Persian)

archers were able to discourage pursuers by firing backwards from the back of a galloping
horse-the famous "Parthian shot."
a. If the speed of the arrow vB the bowstring is vB, that of the horse is vH, the magnitude
of gravitational acceleration is g, and the angle above the horizontal at which the archer
aims is Delta B, what is the range of the shot, measured on the ground from the point of
the shot to the point the arrow falls? Disregard aerodynamic effects and the height
of the horse and rider. Assume the arrow leaves the bow at the angle Delta B at which it is
pointed, as seen by the archer.

b) For vB = 50.0 m/s, vH = 18.0 m/s, and g = 9.81 m/s^2, at what angle Delta B should the
archer aim to maximize the range of part a?

c) For the parameter values of part b, what is the maximum range of the shot, as described
in part a?
d. What is the maximum ground-to-ground range the archer could attain with the same
bow, standing on the ground? That is, how much range does the archer give up for
the Parthian shot? Again, ignore aerodynamic effects and the height of the archer

To solve this problem, we'll use principles of projectile motion. Projectile motion describes the motion of an object in the air under the influence of gravity.

a) To find the range of the shot, we need to determine the horizontal distance traveled by the arrow. We can break down its motion into vertical and horizontal components. Let's assume the time of flight is t.

In the vertical direction, the arrow experiences free fall motion. The equation for the vertical displacement (y) is given by:
y = (1/2) * g * t^2

Since we are disregarding the height of the horse and rider, at the time of impact (when the arrow falls), the vertical displacement is zero. So we can write:
0 = (1/2) * g * t^2

From this equation, we can solve for the time of flight:
t = sqrt((2 * y) / g)
= sqrt(0 / g)
= 0

This indicates that for the arrow to fall at the point of shooting, the time of flight is zero. It means the arrow has no vertical displacement and travels along a horizontal line.

In the horizontal direction, the arrow experiences uniform motion with an initial horizontal velocity (vB) given by the component of the arrow's speed along the horizontal direction.
vB_horizontal = vB * cos(Delta B)

The horizontal displacement (range, R) is given by:
R = vB_horizontal * t
= vB * cos(Delta B) * t
= vB * cos(Delta B) * 0
= 0

So, the horse's speed and the angle at which the archer aims do not affect the horizontal range of the shot in this scenario.

b) To maximize the range of the shot, we need to find the angle (Delta B) at which the horizontal component of the initial velocity is maximized. This occurs when the angle is 45 degrees. So, the archer should aim at an angle of 45 degrees.

c) Since the range is zero as calculated in part a, the maximum range for the given parameters is also zero.

d) If the archer is standing on the ground, the arrow can travel with the same horizontal speed (vB) during its entire flight. In this case, the arrow's range can be calculated by considering the time of flight until it falls to the ground.

Using the equation for the time of flight:
t = sqrt((2 * y) / g)
= sqrt(0 / g)
= 0

Similar to part a, we find that the time of flight is zero, indicating the arrow travels along a horizontal line with no vertical displacement.

Therefore, whether the archer is on horseback or standing on the ground, the range of the shot is zero.