A rocket is launched at an angle of 50.0° above the horizontal with an initial speed of 105 m/s. The rocket moves for 3.00 s along its initial line of motion with an acceleration of 32.0 m/s2. At this time, its engines fail and the rocket proceeds to move as a projectile.

(a) Find the maximum altitude reached by the rocket.
(b) Find its total time of flight.
(c) Find its horizontal range.

To solve this problem, we need to break it down into different parts. Let's start with the first question:

(a) Find the maximum altitude reached by the rocket.

To find the maximum altitude reached, we can use the kinematic equations for projectile motion. The key is to split the rocket's motion into its horizontal and vertical components.

The initial vertical velocity can be found using the equation: v_y = v_0 * sin(θ)
where v_0 is the initial speed of the rocket (105 m/s) and θ is the launch angle (50°).

v_0 = 105 m/s
θ = 50°

v_y = v_0 * sin(θ)
v_y = 105 * sin(50°)
v_y = 80.84 m/s

Next, we need to find the time it takes for the rocket to reach its maximum altitude. We can use the following equation for vertical displacement:

Δy = v_y * t + (1/2) * a_y * t^2

Δy = 0 (maximum height reached)
v_y = 80.84 m/s (initial vertical velocity)
a_y = -9.8 m/s^2 (acceleration due to gravity, negative because it acts in the opposite direction of motion)
t = ? (time taken to reach maximum altitude)

Since the rocket moves vertically under the influence of gravity, the acceleration is -9.8 m/s². The displacement is 0 because the maximum height occurs when the rocket starts falling back down.

Plugging these values into the equation, we get:

0 = 80.84 * t + (1/2) * (-9.8) * t^2

Simplifying this equation, we get a quadratic equation:

-4.9t^2 + 80.84t = 0

Factoring out t, we can solve for t:

t(-4.9t + 80.84) = 0

This equation yields two solutions: t = 0 and t = 16.54 s. Since the rocket took only 3 s to reach its maximum altitude, we discard t = 16.54 s.

Therefore, the maximum altitude reached by the rocket is achieved in 3 s at a vertical velocity of 80.84 m/s:

Δy = v_y * t + (1/2) * a_y * t^2
Δy = 80.84 * 3 + (1/2) * (-9.8) * 3^2
Δy = 122.026 + (-44.145)
Δy = 77.881 m

So, the maximum altitude reached by the rocket is approximately 77.881 m.

Now, let's move on to the second question:

(b) Find its total time of flight.

The total time of flight can be calculated by considering the time it took for the rocket to reach maximum altitude, and then doubling it. Since it took 3 s to reach its maximum altitude, the total time of flight is:

Total time of flight = 2 * time taken to reach maximum altitude
Total time of flight = 2 * 3 s
Total time of flight = 6 s

Therefore, the total time of flight of the rocket is 6 seconds.

Lastly, let's solve the third question:

(c) Find its horizontal range.

The horizontal range is the total horizontal distance covered by the rocket during its flight.

To find the horizontal range, we need to determine how long the rocket spends in motion horizontally.

We know that:
- The initial horizontal velocity (v_x) is given by v_0 * cos(θ).
- The time of flight (T) is 6 s.

Using these values, we can calculate the horizontal range (R):

R = v_x * T
R = v_0 * cos(θ) * T
R = 105 * cos(50°) * 6
R = 105 * 0.642788 * 6
R = 400.72 m

Therefore, the horizontal range of the rocket is approximately 400.72 m.