To determine how long the golf ball is in the air, we can break down the initial velocity into its horizontal and vertical components using trigonometry.
1. Horizontal component: The horizontal component of the initial velocity is given by v_h = v * cos(theta), where v is the magnitude of the initial velocity and theta is the angle above the horizontal. In this case, v = 27 m/s and theta = 30 degrees. Using trigonometric functions, we can calculate the horizontal component as v_h = 27 m/s * cos(30 degrees) = 23.382 m/s.
2. Vertical component: The vertical component of the initial velocity is given by v_v = v * sin(theta). Using the same values of v = 27 m/s and theta = 30 degrees, we can calculate the vertical component as v_v = 27 m/s * sin(30 degrees) = 13.5 m/s.
Now, we can consider the vertical motion of the golf ball. The time it takes for an object to reach its maximum height and fall back down to the same height is the total time the ball is in the air.
3. Time of flight: The time it takes for the ball to reach its maximum height and fall back down is given by the formula t = 2 * (v_v) / g, where g is the acceleration due to gravity (approximately 9.8 m/s^2). Plugging in the values, we have t = 2 * 13.5 m/s / 9.8 m/s^2 = 2.755 seconds.
Therefore, the golf ball is in the air for approximately 2.755 seconds.