1. Understanding the Problem
* Projectile Motion: This is a classic physics problem involving projectile motion. The ball follows a parabolic path due to gravity.
* Initial Velocity Components: The initial velocity (43 m/s) is broken down into two components:
* Horizontal (Vx0): This component remains constant throughout the flight.
* Vertical (Vv0): This component is affected by gravity.
* Time in Air: We want to find the total time the ball spends in the air, from the moment it's thrown until it hits the ground.
2. Solving for the Initial Vertical Velocity (Vv0)
* Trigonometry: We can use trigonometry (SOH CAH TOA) to find Vv0:
* We know the angle (32 degrees) and the hypotenuse (43 m/s).
* Sin(angle) = Opposite / Hypotenuse
* Sin(32°) = Vv0 / 43 m/s
* Vv0 = 43 m/s * Sin(32°) ≈ 22.8 m/s
3. Finding the Time in Air
* Vertical Motion: We'll focus on the vertical motion to find the time.
* Acceleration due to Gravity: The only force acting on the ball vertically is gravity (g ≈ -9.8 m/s²). We use a negative sign since it acts downwards.
* Symmetry: The ball's upward and downward paths are symmetrical. We can find the time it takes to reach the highest point (where Vv = 0) and double it to get the total time in the air.
* Equations of Motion: We'll use the following kinematic equation:
* Vv = Vv0 + at
* Vv = final vertical velocity (0 m/s at the highest point)
* Vv0 = initial vertical velocity (22.8 m/s)
* a = acceleration due to gravity (-9.8 m/s²)
* t = time to reach the highest point
* Solving for t:
* 0 = 22.8 m/s + (-9.8 m/s²) * t
* t ≈ 2.33 seconds
* Total Time in Air:
* Total time = 2 * t ≈ 2 * 2.33 seconds ≈ 4.66 seconds
Therefore:
* The initial vertical velocity component (Vv0) is approximately 22.8 m/s.
* The ball will be in the air for approximately 4.66 seconds.
