A rocket flies upward with a velocity of 1050 km/h and at an angle of 30 degrees to the vertical. If the rocket runs out of fuel determine:

(i) The maximum additional altitude, h.
(ii) The corresponding time.
(iii) The horizontal distance at this time
The gravitational acceleration during this flight is 9.52 m/s^2

a) find vertical velocity: 1050*sin30

mgh=1/2 m (1050sin30)^2 solve for h.
b) h=1050sin30*t-1/2 g t^2 solve for t.
c) distance=1050cos30*t

Thanks a lot

To answer these questions, we need to analyze the motion of the rocket. We will break down the initial velocity into its horizontal and vertical components.

Given:
Initial velocity, v = 1050 km/h
Launch angle, θ = 30 degrees
Gravitational acceleration, g = 9.52 m/s^2

Step 1: Find the initial horizontal and vertical velocities.

The horizontal component of velocity (Vx) remains constant throughout the flight since there is no external force acting in the horizontal direction. Therefore, Vx = v * cos(θ).

Vx = 1050 km/h * cos(30°) = 1050 km/h * (√3/2) ≈ 909.09 km/h.

The vertical component of velocity (Vy) will change due to the acceleration of gravity. Initially, the rocket is moving upward, so Vy = v * sin(θ).

Vy = 1050 km/h * sin(30°) = 1050 km/h * (1/2) = 525 km/h.

Step 2: Determine the maximum additional altitude (h).

At the highest point of the rocket's trajectory, the vertical velocity becomes zero (Vy = 0). We can find the time it takes for Vy to reach zero and calculate the additional altitude.

Using the kinematic equation v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time:

Final vertical velocity (Vfy) = 0
Initial vertical velocity (Vy) = 525 km/h = 525,000 m/3600 s ≈ 145.83 m/s
Acceleration due to gravity (g) = 9.52 m/s^2

Using v = u + at and rearranging, we get t = (Vfy - Vy) / (-g).

t = (0 - 145.83) / (-9.52) ≈ 15.31 s.

Now, we can find the additional altitude (h) using the equation s = ut + (1/2)at^2, where s is the displacement.

Displacement (s) = ?
Initial velocity (u) = 145.83 m/s
Time (t) = 15.31 s
Acceleration (a) = -9.52 m/s^2 (negative because the rocket is moving opposite to the direction of acceleration due to gravity)

s = ut + (1/2)at^2

s = 145.83 * 15.31 + (1/2) * (-9.52) * (15.31)^2 ≈ 1105.22 m

So, the maximum additional altitude (h) is approximately 1105.22 meters.

Step 3: Find the corresponding time.

Since the rocket runs out of fuel, we need to consider the time it takes for the rocket to reach its maximum additional altitude, which we found in the previous step.

The total time of flight will be twice the time taken to reach the maximum altitude (t_total = 2t).

t_total = 2 * 15.31 ≈ 30.63 s

So, the corresponding time is approximately 30.63 seconds.

Step 4: Calculate the horizontal distance traveled at this time.

The horizontal distance (d) traveled during the time of flight can be calculated using the equation d = Vx * t_total.

Horizontal velocity (Vx) = 909.09 km/h = 909.09 * 1000 m/3600 s ≈ 252.52 m/s
Time (t_total) = 30.63 s

d = 252.52 * 30.63 ≈ 7735.73 m

Therefore, the horizontal distance at this time is approximately 7735.73 meters.