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 find the answers to these questions, we will use the principles of projectile motion. We'll break down the problem step by step.
a) The range of the shot can be calculated using the horizontal and vertical components of the initial velocity of the arrow. Since the horse is galloping in a straight line, the horizontal velocity of the arrow is equal to the velocity of the horse, vH. The vertical component of the initial velocity can be found by using the speed of the arrow vB and the angle at which the archer aims, Delta B.
The horizontal distance, or range (R), can be calculated using the formula:
R = (vH * vB * sin(2 * Delta B)) / g
b) To maximize the range of the shot, we need to find the angle at which the range is maximum. The maximum range is achieved when the angle, Delta B, is equal to 45 degrees. This can be proven using calculus, but for simplicity, we can state that it is a general property of projectile motion that the maximum range is achieved when the projectile is launched at a 45-degree angle.
c) To calculate the maximum range, we substitute the values provided into the formula in part a. We have vH = 18.0 m/s, vB = 50.0 m/s, and g = 9.81 m/s^2. Using these numerical values and Delta B = 45 degrees, we can calculate the maximum range.
R(max) = (18.0 * 50.0 * sin(90)) / 9.81
d) To find the maximum ground-to-ground range attainable by the archer standing on the ground, we can consider that the arrow is launched horizontally. In this case, the vertical component of the initial velocity is zero. The horizontal distance covered by the arrow in this case is given by the formula:
R(standing) = (vH^2 / g)
This formula assumes that the arrow is launched horizontally and not at an angle.
Now, you can substitute the values into the appropriate formulas to calculate the answers for parts a, b, c, and d.