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a rocket of mass of 450000kg is in flight. its thrust is directed at angle of 62.5 degree above horizantal n has magnitude of 6000000 N. find the magnitude and direction of rockets acceleration. give direction as an angle above horizontal

To find the magnitude and direction of the rocket's acceleration, we need to use Newton's second law of motion, 𝐹 = π‘šπ‘Ž, where 𝐹 is the net force applied on the rocket, π‘š is its mass, and π‘Ž is its acceleration.

First, we need to resolve the thrust force into its horizontal and vertical components. The horizontal component of the thrust force can be found using πΉβ„Ž = 𝐹 Γ— cos(πœƒ), where πœƒ is the angle above the horizontal.

πΉβ„Ž = 6000000 N Γ— cos(62.5Β°)
= 6000000 N Γ— 0.456 (rounded to three decimal places)
β‰ˆ 2736000 N

The vertical component of the thrust force can be found using 𝐹𝑣 = 𝐹 Γ— sin(πœƒ).

𝐹𝑣 = 6000000 N Γ— sin(62.5Β°)
= 6000000 N Γ— 0.890 (rounded to three decimal places)
β‰ˆ 5340000 N

Now, we can find the net force acting on the rocket in each direction using Newton's second law.

In the horizontal direction, there is no other force acting on the rocket, so the net force is equal to the horizontal component of the thrust force:

𝐹_𝑛𝑒𝑑_β„Ž = πΉβ„Ž
= 2736000 N

In the vertical direction, there is weight acting in the opposite direction. The weight of the rocket (𝑀) can be found using 𝑀 = π‘š Γ— 𝑔, where 𝑔 is the acceleration due to gravity.

𝑀 = 450000 kg Γ— 9.8 m/sΒ²
β‰ˆ 4410000 N

Therefore, the net force in the vertical direction is:

𝐹_𝑛𝑒𝑑_𝑣 = 𝐹𝑣 - 𝑀
= 5340000 N - 4410000 N
= 930000 N

Now, we can find the magnitude and direction of the rocket's acceleration using the net forces in each direction:

π‘Ž = 𝐹/π‘š

In the horizontal direction: π‘Ž_β„Ž = 𝐹_𝑛𝑒𝑑_β„Ž/π‘š
= 2736000 N / 450000 kg
β‰ˆ 6.08 m/sΒ²

In the vertical direction: π‘Ž_𝑣 = 𝐹_𝑛𝑒𝑑_𝑣/π‘š
= 930000 N / 450000 kg
β‰ˆ 2.07 m/sΒ²

The magnitude of the rocket's acceleration is the vector sum of its horizontal and vertical components. Using the Pythagorean theorem, we can calculate:

|π‘Ž| = √(π‘Ž_β„ŽΒ² + π‘Ž_𝑣²)
= √((6.08 m/s²)² + (2.07 m/s²)²)
β‰ˆ √(36.97 + 4.28)
β‰ˆ √41.25
β‰ˆ 6.42 m/sΒ² (rounded to two decimal places)

To find the direction of the rocket's acceleration as an angle above the horizontal (πœƒ), we can use the inverse tangent:

πœƒ = tan⁻¹(π‘Ž_𝑣/π‘Ž_β„Ž)
= tan⁻¹(2.07 m/s² / 6.08 m/s²)
β‰ˆ tan⁻¹(0.34)
β‰ˆ 19Β° (rounded to the nearest degree)

Therefore, the magnitude of the rocket's acceleration is approximately 6.42 m/sΒ², and its direction above the horizontal is approximately 19Β°.