I need help setting up this problem.
A tennis player serves a tennis ball such that it is moving horizontally when it leaves the racquet. When the ball travels a horizontal distance of 19 m, it has dropped 58 cm from its original height when it left the racquet. What was the initial speed of the tennis ball? (Neglect air resistance.)
Thank you.
To set up this problem, we can use the equations of motion for projectile motion. Here's how to do it step by step:
1. Identify the known quantities and what you need to find:
- Known quantities:
- Horizontal distance traveled (d) = 19 m
- Vertical displacement (Δy) = -58 cm (-0.58 m) [negative because the ball dropped]
- Acceleration due to gravity (g) = 9.8 m/s² (assuming the ball is near the Earth's surface)
- Unknown quantity:
- Initial speed of the tennis ball (u)
2. Find the time it takes for the ball to travel the horizontal distance:
- Since the ball is only moving horizontally, there is no acceleration in the horizontal direction (ax = 0).
- Therefore, we can use the equation: d = ut + (1/2)at², where a = 0.
- Substituting the known values: 19 = u * t
- Solve for t: t = 19 / u
3. Use the time found in step 2 to calculate the initial vertical velocity:
- The vertical equation of motion is: Δy = ut + (1/2)at²
- Substituting the known values: -0.58 = u * t + (1/2) * (-9.8) * t²
- Substituting t from step 2: -0.58 = u * (19 / u) + (1/2) * (-9.8) * (19 / u)²
4. Simplify and solve for u:
- Simplifying the equation: -0.58 = 19 - (9.8 * 19²) / (2u)
- Rearranging the equation: 0 = -0.58 * u - 19 + (9.8 * 19²) / (2)
- Solve for u: u = ((9.8 * 19²) / (2)) / 0.58
5. Calculate the value of u:
- Substitute the known values: u = ((9.8 * 19²) / (2)) / 0.58
- Calculate the result: u ≈ 22.73 m/s
Therefore, the initial speed of the tennis ball was approximately 22.73 m/s.