I need help setting up this problem.
While standing on the roof of a building, a child tosses a tennis ball with an initial speed of 15 m/s at an angle of 20° below the horizontal. The ball lands on the ground 2.8 s later. How tall is the building?
Thank you.
To solve this problem, you can use the equations of motion for projectile motion.
Step 1: Analyze the given information
- Initial speed of the ball (vi): 15 m/s
- Launch angle (θ): 20° below the horizontal
- Time of flight (t): 2.8 s
Step 2: Break down the initial velocity into horizontal and vertical components
The initial velocity of the ball (vi) can be broken down into its horizontal and vertical components.
- Horizontal component (vi_x): vi * cos(θ)
- Vertical component (vi_y): vi * sin(θ)
Step 3: Solve for the time of flight (t)
The time of flight (t) is the total time the ball is in the air. In this case, it is given as 2.8 seconds.
Step 4: Solve for the vertical displacement (Δy)
The vertical displacement is the height from the starting point to the landing point of the ball. We can use the equation:
Δy = vi_y * t + 0.5 * g * t^2
where g is the acceleration due to gravity (9.8 m/s^2).
Step 5: Calculate the height of the building
The height of the building is equal to the vertical displacement (Δy).
Now, let's calculate the height of the building.
1. Calculate the horizontal component (vi_x): vi * cos(θ)
vi_x = 15 * cos(20°)
2. Calculate the vertical component (vi_y): vi * sin(θ)
vi_y = 15 * sin(20°)
3. Calculate the vertical displacement (Δy): vi_y * t + 0.5 * g * t^2
Δy = (vi_y * t) + (0.5 * g * t^2)
4. Substitute the given values and calculate Δy.
5. The height of the building is equal to Δy.
By following these steps and using the given information, you can find the height of the building.