A cricketer can throw a ball to a maximum horizontal distance of 100 m. The speed with which he throws the ball is (to the nearest integer)

Options

A. `30 ms`
B. `42ms`
C. `32 ms`
D. `35 ms`

Answer

To find the speed with which the cricketer throws the ball, we can use the formula for the maximum range of projectile motion. The maximum range \( R \) is given by the equation:

\[ R = \frac{u^2 \sin(2\theta)}{g} \]

Where:
- \( R \) is the range (maximum horizontal distance),
- \( u \) is the initial speed,
- \( \theta \) is the angle of projection,
- \( g \) is the acceleration due to gravity (approximately \( 9.8 \, \text{m/s}^2 \)).

### Step 1: Identify the maximum range
From the problem, we know that the maximum horizontal distance \( R \) is 100 m.

### Step 2: Determine the angle for maximum range
The maximum range occurs at an angle of \( 45^\circ \). At this angle, \( \sin(2\theta) = \sin(90^\circ) = 1 \).

### Step 3: Substitute values into the range formula
Substituting the known values into the range formula:

\[ 100 = \frac{u^2 \cdot 1}{9.8} \]

### Step 4: Rearranging the equation to solve for \( u^2 \)
Rearranging gives:

\[ u^2 = 100 \cdot 9.8 \]

### Step 5: Calculate \( u^2 \)
Calculating \( u^2 \):

\[ u^2 = 980 \]

### Step 6: Take the square root to find \( u \)
Now, taking the square root to find \( u \):

\[ u = \sqrt{980} \]

### Step 7: Calculate \( u \)
Calculating \( u \):

\[ u \approx 31.3 \, \text{m/s} \]

### Step 8: Round to the nearest integer
Rounding \( 31.3 \) to the nearest integer gives:

\[ u \approx 31 \, \text{m/s} \]

### Final Answer
The speed with which the cricketer throws the ball is approximately **31 m/s**.

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