A cricket ball is thrown up with a speed of 19.6 m/s. The maximum height it can reach is (take g= 9.8 m/`s^2`)

Options

A. `9.8m`
B. `19.6m`
C. `29.4m`
D. `39.2m`

Answer

To find the maximum height reached by a cricket ball thrown upwards with an initial speed of 19.6 m/s, we can use the equations of motion. Here’s a step-by-step solution:

### Step 1: Identify the known values
- Initial velocity (u) = 19.6 m/s (upwards)
- Final velocity (v) = 0 m/s (at the maximum height)
- Acceleration (a) = -g = -9.8 m/s² (since it acts downwards)

### Step 2: Use the third equation of motion
We can use the third equation of motion, which is:
\[ v^2 = u^2 + 2as \]
Where:
- \( v \) is the final velocity,
- \( u \) is the initial velocity,
- \( a \) is the acceleration,
- \( s \) is the displacement (which we will denote as \( H \) for height).

### Step 3: Substitute the known values into the equation
Substituting the known values into the equation:
\[ 0 = (19.6)^2 + 2(-9.8)H \]

### Step 4: Rearrange the equation
Rearranging the equation gives:
\[ 0 = 384.16 - 19.6H \]
\[ 19.6H = 384.16 \]

### Step 5: Solve for H
Now, divide both sides by 19.6 to find \( H \):
\[ H = \frac{384.16}{19.6} \]
\[ H = 19.6 \text{ m} \]

### Conclusion
The maximum height that the cricket ball can reach is **19.6 meters**.

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