A train is moving on a track at `30ms^-1`. A ball is thrown from it perpendicular to the direction of motion with `30 ms^-1` at `45^@` from horizontal. Find the distance of ball from the point of projection on train to the point where it strikes the ground.

Options

A. 90 m
B. `90(sqrt3)` m
C. 60 m
D. `60 (sqrt3)`m

Explanation

To solve the problem step by step, we will analyze the motion of the ball thrown from the train.

### Step 1: Identify the parameters
- Speed of the train, \( v_t = 30 \, \text{m/s} \)
- Speed of the ball, \( u = 30 \, \text{m/s} \)
- Angle of projection, \( \theta = 45^\circ \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)

### Step 2: Resolve the velocity of the ball into components
Since the ball is thrown at an angle of \( 45^\circ \):
- Horizontal component of the velocity, \( u_x = u \cos(\theta) = 30 \cos(45^\circ) = 30 \times \frac{1}{\sqrt{2}} = 15\sqrt{2} \, \text{m/s} \)
- Vertical component of the velocity, \( u_y = u \sin(\theta) = 30 \sin(45^\circ) = 30 \times \frac{1}{\sqrt{2}} = 15\sqrt{2} \, \text{m/s} \)

### Step 3: Calculate the time of flight
The time of flight \( T \) can be calculated using the formula:
\[ T = \frac{2u_y}{g} \]
Substituting the values:
\[ T = \frac{2 \times 15\sqrt{2}}{10} = \frac{30\sqrt{2}}{10} = 3\sqrt{2} \, \text{s} \]

### Step 4: Calculate the horizontal distance traveled by the ball
The horizontal distance \( R \) (range) can be calculated using:
\[ R = u_x \times T \]
Substituting the values:
\[ R = 15\sqrt{2} \times 3\sqrt{2} = 15 \times 3 \times 2 = 90 \, \text{m} \]

### Step 5: Conclusion
The distance of the ball from the point of projection on the train to the point where it strikes the ground is \( 90 \, \text{m} \).

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