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 determine the range of the Parthian shot and answer the subsequent questions, we can use the principles of projectile motion. Projectile motion refers to the motion of an object that is launched into the air and moves under the influence of gravity alone, assuming no air resistance.
a) To find the range of the Parthian shot, we need to determine the horizontal distance the arrow travels. We can break the initial velocity of the arrow into horizontal and vertical components.
The horizontal component of the initial velocity (v_x) is equal to the product of the velocity of the horse (vH) and the cosine of the angle Delta B:
v_x = vH * cos(Delta B)
The vertical component of the initial velocity (v_y) is equal to the product of the velocity of the bowstring (vB) and the sine of the angle Delta B:
v_y = vB * sin(Delta B)
Since there is no initial vertical velocity, the vertical component of the acceleration is equal to the gravitational acceleration (-g).
Using the equations of motion:
Horizontal distance (range) = (horizontal velocity) * (time of flight)
The time of flight can be found from the vertical component of the motion:
Time of flight = (2 * v_y) / g
Substituting the expressions for v_x and v_y into the horizontal distance equation:
Range = (vH * cos(Delta B)) * (2 * vB * sin(Delta B) / g)
b) To maximize the range, we need to find the value of Delta B that makes the expression for range maximum. We can achieve this by taking the derivative of the range equation with respect to Delta B, setting it equal to zero, and solving for Delta B.
c) To determine the maximum range of the shot, substitute the value of Delta B found in part b into the range equation.
d) To find the maximum ground-to-ground range attainable by the archer standing on the ground, we need to consider the maximum launch angle given by the range equation for a projectile on level ground. This angle is 45 degrees. Substituting this value into the range equation will give us the maximum range achievable in this scenario.
Please provide the values of Delta B, vB, vH, and g to proceed with the calculations and find the answers to the specific numerical questions.