A rocket of mass 4.50 105 kg is in flight. its thrust is directed at an angle of 49.0° above the horizontal and has a magnitude of 6.00 106 N. Find the magnitude and direction of the rockets acceleration. Give the direction as an angle above the horizonal.

Dmd

To find the magnitude and direction of the rocket's acceleration, we can break down the thrust force into its horizontal and vertical components.

First, let's identify the given parameters:
Mass of the rocket (m) = 4.50 x 10^5 kg
Thrust force (F) = 6.00 x 10^6 N
Angle of thrust above the horizontal (θ) = 49.0°

Now, let's find the horizontal (Fx) and vertical (Fy) components of the thrust force:
Fx = F * cos(θ)
Fy = F * sin(θ)

Substituting the values:
Fx = (6.00 x 10^6 N) * cos(49.0°)
Fy = (6.00 x 10^6 N) * sin(49.0°)

Calculating Fx and Fy:
Fx ≈ 4.012 x 10^6 N
Fy ≈ 4.590 x 10^6 N

Now, let's calculate the net force in the horizontal (Fx_net) and vertical (Fy_net) directions:
Fx_net = 0 (since there is no horizontal net force)
Fy_net = Fy - mg

Substituting the values:
Fy_net = (4.590 x 10^6 N) - (4.50 x 10^5 kg) * 9.8 m/s^2
Fy_net = (4.590 x 10^6 N) - (4.41 x 10^6 N) ≈ 0.18 x 10^6 N

Now, we can calculate the magnitude of the rocket's acceleration (a):
a = F_net / m

Substituting the values:
a = (0.18 x 10^6 N) / (4.50 x 10^5 kg)
a ≈ 0.4 m/s^2

The magnitude of the rocket's acceleration is approximately 0.4 m/s^2.

Next, let's find the direction of acceleration as an angle above the horizontal (θ_a):
θ_a = arctan(Fy_net / Fx_net)

Substituting the values:
θ_a = arctan((0.18 x 10^6 N) / (4.012 x 10^6 N))

Calculating θ_a:
θ_a ≈ 25.58°

Therefore, the magnitude of the rocket's acceleration is approximately 0.4 m/s^2, and its direction is approximately 25.58° above the horizontal.