A ball is thrown up with a speed of 15 m/s . How high wiil it go before it begins to fall ? `(g=9.8 m//s^(2)`

Options

A. 11.4 m
B. 15 m
C. 4 m
D. None

Explanation

To find out how high the ball will go before it begins to fall, we can use one of the equations of motion. Here’s a step-by-step solution:

### Step 1: Identify the known values
- Initial velocity (u) = 15 m/s (the speed at which the ball is thrown upwards)
- Final velocity (v) = 0 m/s (the speed at the highest point)
- Acceleration due to gravity (g) = 9.8 m/s² (acting downwards)

### Step 2: Use the equation of motion
We will use the third equation of motion, which relates the initial velocity, final velocity, acceleration, and displacement (height in this case):
\[ v^2 = u^2 - 2gh \]

### Step 3: Rearrange the equation to solve for height (h)
Since we need to find the height (h), we can rearrange the equation:
\[ h = \frac{u^2 - v^2}{2g} \]

### Step 4: Substitute the known values into the equation
Now, substitute the values we have:
- \( u = 15 \, \text{m/s} \)
- \( v = 0 \, \text{m/s} \)
- \( g = 9.8 \, \text{m/s}^2 \)

Substituting these values into the equation:
\[ h = \frac{(15)^2 - (0)^2}{2 \times 9.8} \]

### Step 5: Calculate the height
Calculating the values:
\[ h = \frac{225}{19.6} \]
\[ h \approx 11.48 \, \text{m} \]

### Step 6: Round the answer
We can round this to two decimal places:
\[ h \approx 11.4 \, \text{m} \]

### Final Answer
The maximum height the ball will reach before it begins to fall is approximately **11.4 meters**.

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