A gun is fired from a bunker of 2m deep with an initial velocity of 500m/s

at an angle of pi/6
a) Find the parametric equations of the position of the projectile
b) Find the range of the projectile to the nearest meter.
c) Find the maximum height of the projectile to the nearest meter.

a) To find the parametric equations of the position of the projectile, we will consider the motion in the x and y directions separately.

In the x-direction:
The initial velocity in the x-direction is given by:
V₀x = V₀ * cos(θ)
where V₀ is the initial velocity and θ is the launch angle.

Given:
V₀ = 500 m/s
θ = π/6

Substituting the values, we get:
V₀x = 500 * cos(π/6)
V₀x = 500 * √3/2
V₀x = 250√3 m/s

The equation for the horizontal position (x) as a function of time (t) is:
x = V₀x * t

In the y-direction:
The initial velocity in the y-direction is given by:
V₀y = V₀ * sin(θ)

Given:
V₀ = 500 m/s
θ = π/6

Substituting the values, we get:
V₀y = 500 * sin(π/6)
V₀y = 500 * 1/2
V₀y = 250 m/s

The equation for the vertical position (y) as a function of time (t) is:
y = V₀y * t - (1/2) * g * t²
where g is the acceleration due to gravity (approximately 9.8 m/s²).

The parametric equations of the projectile's position are:
x = 250√3 * t
y = 250 * t - (1/2) * 9.8 * t²

b) To find the range of the projectile, we need to determine the total time it takes for the projectile to reach the ground. We can find this by setting the vertical position (y) equal to zero and solving for t:

250 * t - (1/2) * 9.8 * t² = 0

Simplifying the equation, we get:
t(250 - 4.9t) = 0

The possible solutions are t = 0 and 250 - 4.9t = 0. Ignoring the t = 0 solution, we solve for t in the second equation:

250 - 4.9t = 0
4.9t = 250
t = 250 / 4.9
t ≈ 51.02

The total time of flight is approximately 51.02 seconds.

The range (horizontal distance) of the projectile is given by:
range = x = 250√3 * t

Substituting the value of t, we get:
range = 250√3 * 51.02
range ≈ 21579.6

The range of the projectile is approximately 21580 meters.

c) To find the maximum height of the projectile, we can use the y equation. The maximum height occurs at the vertex of the parabolic path. We can find this by determining the t value at the midpoint of the total time of flight.

The midpoint is t = (0 + 51.02) / 2 = 25.51

Substituting this value into the equation for y, we get:
y = 250 * 25.51 - (1/2) * 9.8 * (25.51)²
y ≈ 3114.5

The maximum height of the projectile is approximately 3115 meters.

a) To find the parametric equations of the position of the projectile, we need to break down the motion of the projectile into its horizontal and vertical components.

Let's assume the projectile starts at the origin (0, 0). The horizontal component of the initial velocity is given by:

Vx = V * cos(angle)

where V is the magnitude of the initial velocity (500 m/s) and angle is the launch angle (pi/6).

Therefore, Vx = 500 * cos(pi/6) = 500 * (√3/2) = 250√3 m/s.

The vertical component of the initial velocity is given by:

Vy = V * sin(angle)

Therefore, Vy = 500 * sin(pi/6) = 500 * (1/2) = 250 m/s.

Now, let's denote time as t. The horizontal position of the projectile can be described by the equation:

x = Vx * t = 250√3 * t

The vertical position of the projectile can be described by the equation:

y = Vy * t - (1/2) * g * t^2

where g is the acceleration due to gravity (approximately 9.8 m/s^2).

Since the bunker is 2m deep, the initial vertical position is -2m. Therefore, the equation becomes:

y = Vy * t - (1/2) * g * t^2 - 2

b) To find the range of the projectile, we need to determine the value of t when the projectile hits the ground. The range is the horizontal position at that time.

When the projectile hits the ground, the vertical position becomes zero (y = 0). Substituting this into the equation for y, we get:

0 = Vy * t - (1/2) * g * t^2 - 2

Rearranging the equation, we get a quadratic equation:

(1/2) * g * t^2 - Vy * t + 2 = 0

Solving this quadratic equation for t, we can find the time when the projectile hits the ground. Since it asked for the answer to the nearest meter, we can use the quadratic formula to approximate the value of t.

c) To find the maximum height of the projectile, we need to determine the peak of the vertical motion. At the peak, the vertical velocity becomes zero (Vy = 0). We can use this condition to find the time at which the projectile reaches its maximum height.

Substituting Vy = 0 into the equation for y, we get:

0 = 0 * t - (1/2) * g * t^2 - 2

Again, rearranging the equation and solving the quadratic equation for t, we can find the time when the projectile reaches its maximum height. Since it asked for the answer to the nearest meter, we can use the quadratic formula to approximate the value of t.