A ball is thrown vertically upward. It has a speed of 10 mls when it has reached one half of its maximum height. How high does the ball rise? (Take g=10mls2)?

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Actually the formula is v^2 -u^2= 2gh. then why did you put a minus sign before 2gh?

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Understanding the Problem
To find the maximum height the ball reaches, we can use the given information about its speed at half of its maximum height.
Given Data
- Initial speed (u) = 10 m/s (at half of maximum height)
- Acceleration due to gravity (g) = 10 m/s²
Key Concepts
- At maximum height, the final speed (v) of the ball will be 0 m/s.
- We can use the kinematic equation: v² = u² + 2as, where:
- v = final velocity (0 m/s at max height)
- u = initial velocity (speed at half max height)
- a = acceleration (which will be negative, -g)
- s = distance (height)
Finding the Maximum Height
1. Calculate the height from half to max height:
- The speed at half the height is 10 m/s.
- Using the equation: 0 = (10)² + 2(-10)(h)
- Rearranging gives: 0 = 100 - 20h
- Thus, 20h = 100, leading to h = 5 m.
2. Determine total height:
- Since 5 m is the distance from half to maximum height, the total maximum height (H) is double that:
- H = 2 * h = 2 * 5 = 10 m.
Final Result
- The maximum height the ball rises to is 10 meters.

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