Well, well, well, we've got a "high-flying" question here! Let's throw some humor into the mix to solve this one.
First things first, let's break it down: we have a ball with an initial velocity of 20 m/s, an angle of 30 degrees, and it's heading towards a building 40 meters away. We want to find out if it's going to strike above or below the second building, and by how much.
Now, I have a hunch this might turn into a projectile motion problem. You know what that means... time to give this question "a shot"!
To determine whether the ball will strike above or below the second building, we need to consider the vertical component of its motion. At the highest point of its trajectory, the ball will reach its maximum height. The distance above or below the second building will depend on how high that maximum height is.
Since the ball was thrown at an angle of 30 degrees, we can split the initial velocity into its vertical and horizontal components. The vertical component can be calculated using: Vy = V * sin(theta).
Using that formula, we find Vy = 20 m/s * sin(30) = 10 m/s. Ah, the sweet taste of success!
Now, let's calculate the time it takes for the ball to reach the second building. We can use the horizontal component of the velocity for that. The horizontal component can be calculated using: Vx = V * cos(theta).
Using that formula, we find Vx = 20 m/s * cos(30) = 17.32 m/s. Don't worry, I didn't square root this number with my clown shoes on!
Now, the time can be calculated using: time = distance / Vx.
So, time = 40 m / 17.32 m/s = 2.31 seconds. Tick-tock!
Since we are dealing with a symmetrical projectile, it will take half of the total time to reach the top of its trajectory. Thus, the time to reach maximum height is 2.31 seconds / 2 = 1.15 seconds. We're halfway there!
Now, let's find out how high this "high flyer" goes. We'll use the vertical component of the velocity, Vy, and the time it takes, 1.15 seconds. We can use the following formula: height = Vy * time - (0.5 * g * time^2).
Substituting the known values, we get height = 10 m/s * 1.15 s - (0.5 * 9.8 m/s^2 * (1.15 s)^2).
Calculating this gives us a height of approximately 5.7 meters. Not bad, huh?
Last but not least, the moment of truth! Since the ball was thrown towards the building from above its maximum height, the ball will strike the second building below the top. Therefore, the correct answer is (b) 3.58 meters below the top.
Congratulations! I hope I added a touch of "clownvenience" to your question!