The coach throws a baseball to a player with an initial speed of `20 m//s` at an angle of `45^@` with the horizontal. At the moment the ball is thrown, the player is 50 m from the coach. At what speed and in what direction must the player run to catch the ball at the same height at which it was released? `(g = 10 m//s^2)`.

Explanation

To solve the problem step by step, we will follow these calculations:

### Step 1: Calculate the time of flight of the baseball.
The formula for the time of flight \( T \) of a projectile is given by:

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

Where:
- \( u = 20 \, \text{m/s} \) (initial speed)
- \( \theta = 45^\circ \)
- \( g = 10 \, \text{m/s}^2 \)

Calculating \( \sin 45^\circ \):

\[ \sin 45^\circ = \frac{\sqrt{2}}{2} \]

Now substituting the values into the formula:

\[ T = \frac{2 \times 20 \times \frac{\sqrt{2}}{2}}{10} = \frac{20\sqrt{2}}{10} = 2\sqrt{2} \, \text{s} \]

### Step 2: Calculate the range of the baseball.
The formula for the range \( R \) of a projectile is given by:

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

Where:
- \( \sin 2\theta = \sin 90^\circ = 1 \)

Now substituting the values:

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

### Step 3: Determine the distance the player needs to cover.
The player is initially 50 m away from the coach. Since the ball travels 40 m, the player needs to cover the difference:

\[ \text{Distance to cover} = 40 \, \text{m} - 50 \, \text{m} = -10 \, \text{m} \]

This means the player needs to run 10 m towards the coach.

### Step 4: Calculate the speed required for the player to catch the ball.
The speed \( v \) required is given by:

\[ v = \frac{S}{T} \]

Where:
- \( S = 10 \, \text{m} \) (the distance the player needs to run)
- \( T = 2\sqrt{2} \, \text{s} \)

Substituting the values:

\[ v = \frac{10}{2\sqrt{2}} = \frac{5}{\sqrt{2}} \, \text{m/s} \]

### Step 5: Determine the direction of the player's run.
Since the player needs to run towards the coach, the direction is forward.

### Final Answer:
The player must run at a speed of \( \frac{5}{\sqrt{2}} \, \text{m/s} \) towards the coach.

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