A ball is projected for maximum range with speed `20 ms^-1`. A boy is located at a distance `25 m` from point of throwing start run to catch the ball at the time when the ball was projected. Find the speed of the boy so that he can catch the ball `(Take g = 10 ms^-1)`.

Explanation

To solve the problem of determining the speed of the boy so that he can catch the ball projected for maximum range, we can follow these steps:

### Step 1: Understand the projectile motion
The ball is projected at a speed of \(20 \, \text{m/s}\) at an angle of \(45^\circ\) for maximum range. The formula for the range \(R\) of a projectile is given by:

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

where \(u\) is the initial speed, \(\theta\) is the angle of projection, and \(g\) is the acceleration due to gravity.

### Step 2: Calculate the range of the projectile
Given:
- \(u = 20 \, \text{m/s}\)
- \(\theta = 45^\circ\)
- \(g = 10 \, \text{m/s}^2\)

First, we calculate \(\sin(2\theta)\):

\[ \sin(90^\circ) = 1 \]

Now substituting the values into the range formula:

\[ R = \frac{(20)^2 \cdot 1}{10} = \frac{400}{10} = 40 \, \text{m} \]

### Step 3: Determine the distance the boy needs to cover
The boy is located \(25 \, \text{m}\) from the point of projection. Since the total range of the ball is \(40 \, \text{m}\), the boy needs to cover:

\[ \text{Distance to cover} = 40 \, \text{m} - 25 \, \text{m} = 15 \, \text{m} \]

### Step 4: Calculate the time of flight of the projectile
The time of flight \(T\) for a projectile is given by:

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

Substituting the values:

\[ T = \frac{2 \cdot 20 \cdot \sin(45^\circ)}{10} \]

Since \(\sin(45^\circ) = \frac{1}{\sqrt{2}}\):

\[ T = \frac{2 \cdot 20 \cdot \frac{1}{\sqrt{2}}}{10} = \frac{40/\sqrt{2}}{10} = \frac{4\sqrt{2}}{2} = 2\sqrt{2} \, \text{s} \]

### Step 5: Calculate the speed of the boy
The speed \(v\) of the boy can be calculated using the formula:

\[ v = \frac{\text{Distance}}{\text{Time}} \]

Substituting the distance the boy needs to cover and the time of flight:

\[ v = \frac{15 \, \text{m}}{2\sqrt{2} \, \text{s}} = \frac{15}{2\sqrt{2}} \, \text{m/s} \]

To simplify:

\[ v = \frac{15 \sqrt{2}}{4} \approx 5.3 \, \text{m/s} \]

### Final Answer
The speed of the boy must be approximately \(5.3 \, \text{m/s}\) to catch the ball.

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