A ball thrown horizontally at 22.3 m/s from the roof of a building lands 31.0 m from the base of the building. How high is the building?

How long did it fly?

went 31 m at 22.3 m/s
t = 31/22.3 = 1.39 seconds

h = (1/2) g t^2
= 4.9 (1.39)^2
= 9.47 meters

To find the height of the building, we can use the equations of motion.

When the ball is thrown horizontally, its initial vertical velocity is zero. The only force acting on it is the force due to gravity, which causes it to fall vertically downwards. The horizontal velocity remains constant throughout the motion.

We are given the horizontal velocity (22.3 m/s) and the horizontal distance traveled (31.0 m). We need to find the vertical distance traveled (height of the building).

To do this, we can use the equation:

𝑑 = 𝑣0𝑡 + 1/2 𝑎𝑡^2

Where:
𝑑 is the vertical distance traveled (height of the building)
𝑣0 is the initial vertical velocity (zero in this case)
𝑡 is the time taken to reach the ground (unknown)
𝑎 is the acceleration due to gravity (-9.8 m/s^2)

Since the ball is thrown horizontally, the time taken to reach the ground will be the same as the time it takes to cover the horizontal distance (31.0 m) at the horizontal velocity (22.3 m/s).

Using the equation:

𝑡 = 𝑑/𝑣

Where:
𝑡 is the time taken to cover the distance (31.0 m)
𝑑 is the total horizontal distance (31.0 m)
𝑣 is the horizontal velocity (22.3 m/s)

Substituting the values into the equation:

𝑡 = 31.0 m / 22.3 m/s = 1.39 s (approximately)

Now we can use the equation of motion for vertical distance traveled:

𝑑 = 𝑣0𝑡 + 1/2 𝑎𝑡^2

Substituting the values into the equation:

𝑑 = 0 + 1/2 (-9.8 m/s^2) (1.39 s)^2 = -9.8 m/s^2 * 1.93 s^2 ≈ -18.85 m

Since the distance cannot be negative in this context, we take the absolute value:

𝑑 = 18.85 m (approximately)

Therefore, the height of the building is approximately 18.85 meters.