To determine the maximum height reached by the ball and its velocity when it reaches a height of 25m, we can use the equations of motion for vertical projectile motion.
- Maximum height: In vertical projectile motion, the initial vertical velocity (u) is given as 25 m/s, and the final vertical velocity (v) at the maximum height is 0 m/s. The acceleration due to gravity (g) is approximately 9.8 m/s² (assuming no air resistance). We need to find the displacement (s) or the maximum height reached by the ball.
Using the equation of motion: v² = u² + 2as
Substituting the known values: 0² = 25² + 2(-9.8)s
Simplifying the equation: 0 = 625 - 19.6s
19.6s = 625
s = 625 / 19.6 s ≈ 31.89 meters
Therefore, the ball reaches a maximum height of approximately 31.89 meters.
- Velocity at a height of 25m: Now we need to find the velocity of the ball when it reaches a height of 25 meters. Let's denote the final velocity as vf. We can use the equation of motion: v² = u² + 2as
Substituting the known values: vf² = 25² + 2(-9.8)(25)
Simplifying the equation: vf² = 625 - 490 vf² = 135 vf ≈ √135 vf ≈ 11.62 m/s
Therefore, the velocity of the ball when it reaches a height of 25 meters is approximately 11.62 m/s.