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You launch a water balloon from the ground with a speed of 7.3 m/s at an angle of 25 degrees. What is the horizontal component of the velocity?

User Indira
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Final answer:

The horizontal component of the velocity for a water balloon launched at 7.3 m/s at a 25-degree angle is approximately 6.63 m/s, calculated using the formula Vx = V * cos(θ).

Step-by-step explanation:

The question involves calculating the horizontal component of the velocity for a water balloon launched at a certain angle and initial speed, which is a topic in projectile motion. To find the horizontal component (Vx) of the velocity, you can use the formula Vx = V * cos(θ), where V is the initial velocity and θ is the launch angle.

In this case, the initial velocity (V) is given as 7.3 m/s and the angle (θ) is 25 degrees. Plugging these values into the formula gives us:

Vx = 7.3 m/s * cos(25°) = 6.63 m/s (approximately)

Therefore, the horizontal component of the velocity for the water balloon is approximately 6.63 m/s.

User DIXONJWDD
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