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What is the horizontal component of a rocket launched at a speed of 1.5x10³ m/s with an angle of 68°?

A) 6.59x10² m/s
B) 7.21x10² m/s
C) 8.04x10² m/s
D) 9.12x10² m/s

User Jkelley
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1 Answer

7 votes

Final answer:

The horizontal component of a rocket's velocity launched at a speed of 1.5x10³ m/s at an angle of 68° is approximately 561.9 m/s. Hence, the closest given answer is A) 6.59x10² m/s.

Step-by-step explanation:

The horizontal component of a rocket's velocity can be determined using trigonometric functions. Specifically, the horizontal component Vx is found by multiplying the rocket's launch speed by the cosine of the launch angle. In this case, the launch speed is 1.5x10³ m/s and the launch angle is 68°.

To find the horizontal component of the velocity:

Vx = V * cos(θ)

Vx = 1.5x10³ m/s * cos(68°)

Vx = 1.5x10³ m/s * 0.3746 (cosine of 68 degrees)

Vx ≈ 561.9 m/s

Therefore, the closest choice to the calculated value is A) 6.59x10² m/s.

User Ivo Bosticky
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