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 equations of motion and the principles of projectile motion. Let's break down each part of the problem and solve it step by step:
a) To find the range of the shot, we need to determine the horizontal distance traveled by the arrow.
The horizontal component of the arrow's initial velocity (vB) will determine the range. We can find this component using trigonometry. The horizontal component of the velocity (vB_x) is given by vB * cos(Delta B).
The time of flight (t) of the arrow can be found using the equation t = 2 * vB * sin(Delta B) / g.
Finally, we can calculate the horizontal range (R) using the equation R = vB_x * t.
b) To maximize the range (R), we need to find the angle (Delta B) that gives the maximum value for R. We can achieve this by finding the maximum point of the range function.
To do this, we can take the derivative of the range function with respect to Delta B and set it equal to zero. After finding Delta B, we need to check if it corresponds to a maximum by taking the second derivative of the range function and ensuring it is negative.
c) Once we have found the angle (Delta B) that maximizes the range, we can substitute the given values (vB, vH, g) into the range equation to calculate the maximum range for those parameters.
d) To find the maximum ground-to-ground range for the archer standing on the ground, we can follow the same process as in part a), but this time the initial velocity (vB) will be different.
Let's apply these steps to solve the problem:
a) Given the variables vB, vH, g, and Delta B, we can calculate the range R using the following equations:
vB_x = vB * cos(Delta B)
t = 2 * vB * sin(Delta B) / g
R = vB_x * t
b) To maximize the range, we need to find the angle Delta B that gives the maximum value. To do this, we need to find the derivative of the range equation, set it equal to zero, and solve for Delta B.
dR/d(Delta B) = 0
(2 * vB^2 * sin(Delta B) * cos(Delta B)) / g - (2 * vB^2 * sin(Delta B) * cos(Delta B)) / g = 0
2 * vB^2 * sin(Delta B) * cos(Delta B) = 0
From this equation, we can see that Delta B = 45 degrees is a valid solution.
c) Substitute the given values into the range equation:
vB_x = vB * cos(Delta B)
t = 2 * vB * sin(Delta B) / g
R = vB_x * t
Plug in vB = 50.0 m/s, Delta B = 45 degrees, and g = 9.81 m/s^2 to calculate R.
d) To find the maximum ground-to-ground range for the archer standing on the ground, we can repeat the calculations in part a), but this time considering that the archer is not on a horse. So the variables vH and Delta B will not be used.
I hope this explanation helps you understand how to approach and solve this problem.