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?
Check the answer given by Damon later to the same question.
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To find the height of the building, we need to analyze the motion of the ball. Since the ball is thrown horizontally, we can conclude that the initial vertical velocity (Vy) is 0 m/s. The only force acting on the ball in the vertical direction is gravity.
We can use the equation of motion for vertical motion:
h = Vy*t + (1/2)*g*t^2
Where:
- h represents the height of the building.
- Vy is the initial vertical velocity of the ball.
- g is the acceleration due to gravity (approximately 9.8 m/s^2).
- t represents the time of flight of the ball.
To determine the time of flight (t), we need to use the horizontal velocity (Vx) of the ball, which remains constant throughout its motion. We can calculate the time of flight using the equation:
t = d / Vx
Where:
- d is the horizontal distance the ball travels (31.0 m).
- Vx is the initial horizontal velocity of the ball (22.3 m/s).
Substituting the given values into the equation, we have:
t = 31.0 m / 22.3 m/s
Now, let's find the time of flight (t):
t = 1.39 s (rounded to two decimal places)
Now that we have the time of flight, we can calculate the height of the building (h). Since Vy = 0 m/s, the equation simplifies to:
h = (1/2)*g*t^2
Plug in the known values:
h = (1/2) * 9.8 m/s^2 * (1.39 s)^2
Simplifying the equation, we get:
h = 9.81 m (rounded to two decimal places)
Therefore, the height of the building is approximately 9.81 meters.