Final answer:
The ball remained in the air for approximately 2.47 seconds and had a speed of approximately 24.2 m/s just before striking the ground.
Step-by-step explanation:
To determine the time the ball remained in the air, we can use the equation:
h = 0.5 * g * t^2
Where h is the height, g is the acceleration due to gravity (which is approximately 9.8 m/s^2), and t is the time the ball remained in the air.
Using the given height, we can rearrange the equation to solve for t:
t = √(2 * h / g)
Plugging in the values, we have t = √(2 * 30.0m / 9.8 m/s^2) = √6.12 s ≈ 2.47 s.
To find the ball's speed just before striking the ground, we can use another equation:
v^2 = u^2 + 2 * g * h
Where v is the final velocity, u is the initial velocity (which in this case is 0 m/s since the ball was dropped), g is the acceleration due to gravity, and h is the height. Rearranging the equation, we have:
v = √(u^2 + 2 * g * h)
Plugging in the values, we have v = √(0 + 2 * 9.8 m/s^2 * 30.0m) = √588 m^2/s^2 ≈ 24.2 m/s.
Learn more about the time the ball remained in the air and the ball's speed before striking the ground