A baseball is hit with a speed of 27.0 at an angle of 48.0. It lands on the flat roof of a 12.0 -tall nearby building.If the ball was hit when it was 1.0 above the ground, what horizontal distance does it travel before it lands on the building?
To find the horizontal distance that the baseball travels before it lands on the building, we can use the equations of motion for projectile motion.
First, let's analyze the given information:
- Initial velocity (v₀) = 27.0 m/s
- Launch angle (θ) = 48.0°
- Initial height (h₀) = 1.0 m
- Height of the building (h) = 12.0 m
We need to find the horizontal distance (x) traveled by the baseball.
To solve this problem, we can follow these steps:
Step 1: Break down the initial velocity into its horizontal and vertical components.
- The horizontal component of velocity (v₀x) is given by: v₀x = v₀ * cos(θ)
- The vertical component of velocity (v₀y) is given by: v₀y = v₀ * sin(θ)
Step 2: Calculate the time of flight (t) for the baseball to reach the building.
- The time of flight can be obtained using the following equation: t = (2 * v₀y) / g
where g is the acceleration due to gravity, which is approximately 9.8 m/s².
Step 3: Calculate the horizontal distance traveled by the baseball.
- The horizontal distance can be determined by multiplying the horizontal component of velocity by the time of flight: x = v₀x * t
Let's plug in the given values and calculate the answer:
Step 1:
- v₀x = 27.0 m/s * cos(48.0°)
- v₀y = 27.0 m/s * sin(48.0°)
Step 2:
- t = (2 * v₀y) / g
- t = (2 * 27.0 m/s * sin(48.0°)) / 9.8 m/s²
Step 3:
- x = v₀x * t
Now, you can perform the calculations to find the horizontal distance traveled by the baseball before it lands on the building.