You are launching Robin Hood out of a cannon and over the wall of the castle to rescue the Maid Marion. He would like to clear the castle wall by as much as possible. The cannon is stuck at an angle of 45 degrees from the ground and will launch him with a speed of 25 m/s. How close does he need to be to the wall to have the largest gap between him and the top of the wall as he flies over it?

clearly he wants the vertex of the parabola to be just above the wall.

Recall that the trajectory can be described as

y = x tanθ - g/[2 v^2 cos^2θ] x^2

Plugging in your values, that is

y = 0.707x - 0.196x^2

The vertex is at (1.8,0.6)

So, a distance of 1.8 meters will provide the maximum clearance.

To determine how close Robin Hood needs to be to the wall to have the largest gap between him and the top of the wall as he flies over it, we can use projectile motion equations.

Step 1: Analyze the problem.
We know the launch angle is 45 degrees, the launch speed is 25 m/s, and we need to find the distance from the wall.

Step 2: Break the initial velocity into horizontal and vertical components.
The horizontal component (Vx) remains constant throughout the motion, while the vertical component (Vy) changes due to the acceleration of gravity.

Vx = V * cosθ (where θ is the launch angle)
Vx = 25 m/s * cos(45°)
Vx = 25 m/s * √2 / 2
Vx = 25 m/s * 0.707
Vx ≈ 17.68 m/s

Vy = V * sinθ (where θ is the launch angle)
Vy = 25 m/s * sin(45°)
Vy = 25 m/s * √2 / 2
Vy = 25 m/s * 0.707
Vy ≈ 17.68 m/s

Step 3: Determine the time it takes to reach the highest point.
The time it takes for the projectile to reach its maximum height is the same as the time it takes to fall to the highest point.

At the maximum height, the vertical component of the velocity becomes zero.
Using the equation Vy = Vy0 - g * t, and substituting Vy = 0, Vy0 = 17.68 m/s, and g = 9.8 m/s²:
0 = 17.68 m/s - 9.8 m/s² * t_max
t_max ≈ 17.68 m/s / 9.8 m/s²
t_max ≈ 1.80 seconds

Step 4: Calculate the maximum height reached.
Using the equation for vertical displacement at time t, with Vy0 = 17.68 m/s and t = t_max:
y_max = Vy0 * t_max - (1/2) * g * t_max²
y_max ≈ 17.68 m/s * 1.80 s - (1/2) * 9.8 m/s² * (1.80 s)²
y_max ≈ 31.81 m - 15.77 m
y_max ≈ 16.04 meters

Step 5: Determine the range or horizontal distance traveled.
The horizontal distance traveled can be calculated using the equation x = Vx * t_total, where t_total is the total time of flight.

The total time of flight is twice the time to reach the maximum height (2 * t_max).
t_total = 2 * t_max
t_total ≈ 2 * 1.80 s
t_total ≈ 3.60 seconds

Using x = Vx * t_total and substituting Vx = 17.68 m/s and t_total = 3.60 s:
x = 17.68 m/s * 3.60 s
x ≈ 63.65 meters

Step 6: Determine how close Robin Hood needs to be to the wall.
Since the range is the horizontal distance, to have the largest gap between Robin Hood and the top of the wall, he should be positioned exactly at the range distance from the wall.

Therefore, Robin Hood needs to be approximately 63.65 meters away from the wall.

Keep in mind that these calculations assume ideal conditions and neglect air resistance. In a real-world scenario, there might be other factors to consider.