a ball is thrown at an initial angle of 37 and initial velocity of 23.0 m/s reaches a maximum height h, as shown in the Figure. With what initial speed must a ball be thrown straight up to reach the same maximum height h?

90

13.8 To find the initial speed required to reach the same maximum height, we can use the principle of conservation of mechanical energy. Assuming no air resistance, we know that the total mechanical energy remains constant throughout the trajectory, which means the potential energy at the maximum height will be the same for both cases.

1. Find the maximum height reached by the ball thrown at an initial angle:
- Let's consider the vertical motion of the ball.
- The height can be determined using the following equation: Δh = (v₁² * sin²θ) / (2 * g), where v₁ is the initial velocity, θ is the launch angle, and g is the acceleration due to gravity (approximately 9.8 m/s²).
- Substituting the given values:
Δh = (23.0 m/s)² * sin²(37°) / (2 * 9.8 m/s²)
≈ 8.92 meters

2. Find the initial speed required for a ball thrown straight up to reach the same maximum height:
- When the ball is thrown straight up, the launch angle is 90 degrees.
- The maximum height can be determined using the same equation as before.
- Setting the height to be 8.92 meters and θ to be 90 degrees, we can solve the equation for v.
8.92 m = (v² * sin²(90°)) / (2 * 9.8 m/s²)
v² = 8.92 m * 2 * 9.8 m/s² / sin²(90°)
v = √(8.92 m * 2 * 9.8 m/s² / 1)
≈ 9.85 m/s

Therefore, to reach the same maximum height, the ball must be thrown straight up with an initial speed of approximately 9.85 m/s.