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 can use the principles of projectile motion. We'll break the problem down into parts and solve them step by step.

a) The range of the shot can be calculated using the horizontal velocity of the arrow and the time of flight. The horizontal velocity can be determined from the speed of the arrow and the angle at which the archer aims.

Let's assume the range of the shot is R. The horizontal velocity of the arrow (vAx) can be found using the following equation:

vAx = vB * cos(Delta B)

Here, vB is the speed of the arrow, and Delta B is the angle above the horizontal at which the archer aims.

The time of flight of the arrow (t) can be calculated using the following equation of motion:

R = vAx * t

Now, to find the vertical velocity (vAy) of the arrow when it hits the ground, we can use the equation of motion for vertical motion under constant acceleration:

0 = vAy - g * t

Rearranging this equation, we get:

t = vAy / g

Substituting this value in the equation for R:

R = vAx * (vAy / g)

Now, we need to find the vertical velocity of the arrow at the time of impact. We can assume that the arrow hits the ground at the same height at which it was released. So we have:

vAy = vB * sin(Delta B)

Substituting this value in the equation for R:

R = vB * cos(Delta B) * (vB * sin(Delta B) / g)

Simplifying:

R = (vB^2 / g) * cos(Delta B) * sin(Delta B)

This gives us the range of the shot.

b) To maximize the range, we need to find the angle Delta B that gives the maximum value for R. To do this, we can differentiate the range equation with respect to Delta B and set it equal to zero:

dR/d(Delta B) = 0

Differentiating the equation for R, we get:

dR/d(Delta B) = (vB^2 / g) * (cos^2(Delta B) - sin^2(Delta B))

Setting this equal to zero:

cos^2(Delta B) - sin^2(Delta B) = 0

Using the trigonometric identity cos^2(A) - sin^2(A) = cos(2A), we have:

cos(2 * Delta B) = 0

This implies that 2 * Delta B = 90 degrees or 270 degrees.

So, the maximum range occurs when the angle Delta B is either 45 degrees or 135 degrees.

c) To find the maximum range, we substitute the values of vB, vH, and g into the range equation:

R = (vB^2 / g) * cos(Delta B) * sin(Delta B)

Using Delta B = 45 degrees and Delta B = 135 degrees, we calculate the respective ranges of the shot.

d) To find the maximum ground-to-ground range attainable by the archer standing on the ground, we can use similar principles of projectile motion but without considering the horse's speed.

You can follow the same steps as in part a but assume vH = 0. This would give you the maximum range achievable by the archer standing on the ground.

Keep in mind that these calculations do not account for aerodynamic effects or other factors like the height of the archer or the horse.