You are a Middle Ages cannonball scientist attacking a castle a distance D away. Your job is to get a cannonball of mass m over the walls, which are a height H above the top of the cannonball launcher. At the instant when the cannonball leaves the launcher it will be traveling at a speed v0 directed at an angle ¦È above the horizontal.
1. Write the cannonball¡¯s coordinates x(t) and y(t) in terms of v0, ¦È, m, and g, or a subset of these quantities. Define your x-y coordinate system as shown in the figure with the origin at the end of the cannonball launcher, and set the time to t = 0 at the instant when the cannonball leaves the launcher.
2. At what time tD, does the x-coordinate of the cannonball reach the x-coordinate of the wall?
3. What is the value of the y-coordinate of the cannonball at the instant when the x-coordinate reaches the castle wall?
1. To write the cannonball's coordinates x(t) and y(t), we can use the equations of projectile motion. Let's break down the motion into its horizontal and vertical components.
The horizontal position x(t) can be determined using the equation:
x(t) = v0 * cos(φ) * t
Here, v0 is the initial velocity of the cannonball, φ is the angle at which it is launched, and t is the time since the cannonball was launched. We are assuming no air resistance.
The vertical position y(t) can be determined using the equation for vertical motion under gravity:
y(t) = H + v0 * sin(φ) * t - (1/2) * g * t^2
Here, H is the height of the castle wall above the cannonball launcher, g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time since the cannonball was launched.
2. To find the time tD when the x-coordinate of the cannonball reaches the x-coordinate of the wall, we equate the two values. Let's assume the distance from the cannonball launcher to the castle wall is D:
x(tD) = D
Substituting the equation for x(t) from Step 1:
v0 * cos(φ) * tD = D
Solving for tD:
tD = D / (v0 * cos(φ))
This gives us the time it takes for the cannonball to reach the horizontal position of the castle wall.
3. To find the value of the y-coordinate of the cannonball when it reaches the castle wall, we substitute tD into the equation for y(t):
y(tD) = H + v0 * sin(φ) * tD - (1/2) * g * tD^2
Substituting the expression for tD from Step 2:
y(tD) = H + (v0 * sin(φ) / (v0 * cos(φ))) * D - (1/2) * g * (D / (v0 * cos(φ)))^2
Simplifying further:
y(tD) = H + (D * tan(φ)) - (1/2) * g * (D^2 / (v0^2 * cos^2(φ)))
This gives us the value of the y-coordinate of the cannonball at the instant when the x-coordinate reaches the castle wall.