Question

A weather team releases a weather balloon. The balloons buoyancy accelerates it straight up at 15 m/s². A wind accelerates it horizontally at 9 m/s². What is the magnitude and direction of the resultant acceleration?

Answer 1

The magnitude of the resultant acceleration is approximately 17.49 m/s², and the direction is 30.96° from the vertical.

Step-by-step explanation:

A weather team releases a weather balloon, which accelerates straight up with a buoyancy of 15 m/s² and is simultaneously pushed horizontally by the wind at 9 m/s². To calculate the magnitude and direction of the resultant acceleration, we treat the problem as a right-angled triangle with the vertical and horizontal accelerations as the perpendicular sides.

Using the Pythagorean theorem:

• Magnitude of resultant acceleration (a) = √((15 m/s²)² + (9 m/s²)²) = √(225 + 81) = √306 = 17.49 m/s² (approximately)

To find the direction, we use the tangent function:

• θ = tan⁻¹(opposite/adjacent) = tan⁻¹(9/15) = 30.96°

Therefore, the resultant acceleration is approximately 17.49 m/s² at an angle of 30.96° from the vertical.