A ball is throw vertically upward. It has a speed of `10 m//s` when it has reached one half of its maximum height. How high does the ball rise ? (Taking `g = 10 m//s^2`).

Options

A. 15 m
B. 10 m
C. 20 m
D. 5 m

Explanation

To solve the problem of how high the ball rises when it is thrown vertically upward, we can follow these steps:

### Step 1: Understand the Problem
We know that the ball has a speed of `10 m/s` when it reaches half of its maximum height. We need to find the maximum height (h) the ball reaches.

### Step 2: Use the Kinematic Equation
We will use the kinematic equation that relates initial velocity (u), final velocity (v), acceleration (a), and displacement (s):
\[ v^2 = u^2 + 2as \]

### Step 3: Set Up the Variables
- At maximum height, the final velocity (v) is `0 m/s`.
- The initial velocity (u) when the ball is at half the maximum height is `10 m/s`.
- The acceleration (a) due to gravity is `-g = -10 m/s²` (negative because it acts downward).
- The displacement (s) when the ball is at half the maximum height is `h/2`.

### Step 4: Apply the Kinematic Equation at Maximum Height
Using the kinematic equation:
\[ 0 = (10)^2 + 2(-10)(h/2) \]

### Step 5: Simplify the Equation
This simplifies to:
\[ 0 = 100 - 10h \]

### Step 6: Solve for Maximum Height (h)
Rearranging gives:
\[ 10h = 100 \]
\[ h = \frac{100}{10} = 10 \text{ meters} \]

### Conclusion
The maximum height the ball rises to is **10 meters**.

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