You are an artillery offcer attached to Gaius Julius Caesar's Xth Legion in the Gallic War
(1st C. BCE). You are responsible for a catapult, a device with a long spoonlike arm used
to lob stones or pumpkins over the walls of Gallic oppida. These fortresses are protected by
high walls rising from a deep, wide moat; you must fire your missiles from the far side of the
moat. You are expected to fire forward.
a) Caesar has asked you for an algebraic formula for the maximum height of wall you
can clear from across a moat of width x, if the initial speed of your projectile is v0,
the magnitude of the acceleration of gravity is g, and you can launch at any angle you
choose. What formula do you give him?* Assume aerodynamic forces are negligible
and disregard the height of the catapult itself.
b) For a certain value of x your formula gives zero height. Explain why-to what does
this correspond?
c)When you shoot to clear a wall of maximum height per the formula of part a, is your
missile ascending, descending, or at the peak of its trajectory when it clears the wall?
a) The formula for the maximum height of the wall you can clear from across a moat of width x, with an initial speed of v0 and gravity acceleration of g, can be derived using the principles of projectile motion. We can use the kinematic equations to determine the maximum height.
Let's assume the launch angle of the projectile is θ, and the time it takes to reach the other side of the moat is t. The horizontal distance (range) traveled by the projectile is equal to the width of the moat, which is x.
Using the equation for horizontal distance: x = v0 * cos(θ) * t
Since we want to find the maximum height, we need to find the time it takes for the projectile to reach that height. We know that at the maximum height, the vertical velocity (vy) will be zero.
Using the equation for vertical velocity: vy = v0 * sin(θ) - g * t
Setting vy equal to zero and solving for t, we get: t = v0 * sin(θ) / g
Substituting this value of t into the equation for horizontal distance, we have: x = v0 * cos(θ) * (v0 * sin(θ) / g)
Simplifying, we get: x = (v0^2 * sin(2θ)) / g
Rearranging the equation to solve for the maximum height (h), we get: h = (v0^2 * sin^2(θ)) / (2g)
So, the algebraic formula for the maximum height of the wall you can clear is h = (v0^2 * sin^2(θ)) / (2g).
b) If the formula gives a zero height for a certain value of x, it means that there is no launch angle θ that would allow you to clear a wall of that particular width (x). In this case, it corresponds to the situation where the projectile would not reach a height above the wall but instead hit the wall directly. Therefore, it would be impossible to clear a wall with that specific width using the given initial speed and gravity.
c) When you shoot to clear a wall of maximum height per the formula in part a, your missile is at the peak of its trajectory when it clears the wall. At the peak of its trajectory, the vertical velocity of the projectile becomes zero before it starts descending. This means that the missile has reached its maximum height and is about to start descending when it clears the wall.