A projectile is fired with a velocity u at the entrance A to a horizontal tunnel of length L and height H. Determine the minimum value of u and the corresponding value of the angle θ for which the projectile will reach B at the other end of the tunnel without touching the top of the tunnel.

To determine the minimum value of u and the corresponding angle θ for which the projectile will reach point B without touching the top of the tunnel, we need to use the concept of projectile motion.

Let's break down the problem into two components: horizontal and vertical motion.

Horizontal Motion:
Since the projectile is fired horizontally at the entrance A, the horizontal velocity (Vx) remains constant throughout its motion. Therefore, the time taken to travel the length of the tunnel (L) is given by t = L / Vx.

Vertical Motion:
The vertical motion of the projectile can be described by the equation: y = uy * t - (1/2) * g * t², where uy is the initial vertical velocity, g is the acceleration due to gravity (approximately 9.8 m/s²), and y is the height.

We need to find the minimum value of u and the corresponding angle θ so that the projectile just reaches the height of the tunnel H at the end B. Therefore, we set y = H, and solve for uy.

H = uy * t - (1/2) * g * t²
H = (u * sinθ) * (L / (u * cosθ)) - (1/2) * g * (L / (u * cosθ))²
H = L * tanθ - (1/2) * g * (L² / (u² * cos²θ))

Simplifying the equation and rearranging, we get:
u² * cos²θ = (L * (L * tanθ - 2H)) / g

Now, we need to minimize the value of u, which means we need to maximize the denominator on the right-hand side of the equation. For that, we set the derivative of the denominator with respect to θ equal to zero.

d/dθ [(L * (L * tanθ - 2H)) / g] = 0

Simplifying this equation and solving for θ will give us the angle at which the derivative is zero, which corresponds to the maximum height of the projectile without touching the top of the tunnel.

Lastly, to find the minimum value of u, we substitute the value of θ obtained and solve for u using the equation:
u² * cos²θ = (L * (L * tanθ - 2H)) / g

By solving these equations, you can determine the minimum value of u and the corresponding angle θ for the projectile to reach point B without touching the top of the tunnel.

To determine the minimum value of u and the corresponding value of the angle θ for which the projectile will reach point B without touching the top of the tunnel, we can use the equations of projectile motion.

Let's break down the problem step-by-step:

Step 1: Set up the coordinate system
- Choose a coordinate system where the x-axis is horizontal and the y-axis is vertical.
- Place the origin at point A, the entrance of the tunnel.

Step 2: Analyze the projectile's motion in the x-axis
- The projectile's initial velocity along the x-axis is u*cos(θ), where θ is the angle of projection.
- The distance from A to B is L.

Using the equation of motion: x = v0x * t, where v0x = u*cos(θ) and x = L, we can find the time of flight (t).
L = u*cos(θ) * t
t = L / (u*cos(θ))

Step 3: Analyze the projectile's motion in the y-axis
- The projectile's initial velocity along the y-axis is u*sin(θ).
- The projectile's vertical motion is affected by gravity.

The maximum height (ymax) reached by the projectile can be determined using the equation: ymax = (u*sin(θ))^2 / (2g),
where g is the acceleration due to gravity.

The projectile will hit the top of the tunnel if ymax > H.

Step 4: Solve for the minimum value of u and the corresponding value of θ
- Set ymax = H and solve for u:

H = (u*sin(θ))^2 / (2g)
u*sin(θ) = sqrt(2gH)
u = sqrt(2gH) / sin(θ)

Plug this value of u into the equation for time of flight (t) from step 2:

t = L / (u*cos(θ))
t = L / ((sqrt(2gH) / sin(θ)) * cos(θ))
t = L * sin(θ) / (sqrt(2gH) * cos(θ))

Now we have an expression for the time of flight (t) in terms of θ. To find the minimum value of u and the corresponding value of θ, we need to minimize t.

Differentiate t with respect to θ and set the derivative equal to zero:

d/dθ (L * sin(θ) / (sqrt(2gH) * cos(θ))) = 0
(L * cos(θ) * sqrt(2gH) - L * sin(θ) * sqrt(2gH)) / (sqrt(2gH) * cos²(θ)) = 0
L * cos(θ) - L * sin(θ) = 0
cos(θ) = sin(θ)

Since cos(θ) = sin(θ) at θ = π/4 (45 degrees), this is the critical angle that minimizes t.

Substitute θ = π/4 into the equations for t and u:

t = L * sin(π/4) / (sqrt(2gH) * cos(π/4))
u = sqrt(2gH) / sin(π/4)

Simplify the equations:
t = L / (sqrt(2gH) / sqrt(2))
u = sqrt(2gH)

Therefore, the minimum value of u is sqrt(2gH), and the corresponding value of θ is π/4 (45 degrees).

There must be a diagram that goes with this.