A particle of mass 50 g is projected horizontally from the top of a tower of height 50 m with a velocity `20m//s`. If `g=10m//s^(2)`, then find the :-

Options

A. Velocity at the instant t=2 s
B. Position at the instant t=3 s
C. Velocity just before hitting the ground
D. Change in velocity in 2 s

Answer

To solve the problem step by step, we will break it down into parts as follows:

### Given Data:
- Mass of the particle, \( m = 50 \, \text{g} = 0.05 \, \text{kg} \) (conversion from grams to kilograms)
- Height of the tower, \( h = 50 \, \text{m} \)
- Initial horizontal velocity, \( u_x = 20 \, \text{m/s} \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)

### Part 1: Velocity at \( t = 2 \, \text{s} \)

1. **Horizontal Component of Velocity:**
- The horizontal component of velocity remains constant.
\[ v_x = u_x = 20 \, \text{m/s} \]

2. **Vertical Component of Velocity:**
- The vertical component of velocity can be calculated using the equation:
\[ v_y = u_y + g \cdot t \]
where \( u_y = 0 \) (initial vertical velocity).
\[ v_y = 0 + 10 \cdot 2 = 20 \, \text{m/s} \]

3. **Magnitude of the Overall Velocity:**
- The overall velocity \( v \) can be calculated using the Pythagorean theorem:
\[ v = \sqrt{v_x^2 + v_y^2} = \sqrt{(20)^2 + (20)^2} = \sqrt{400 + 400} = \sqrt{800} = 20\sqrt{2} \, \text{m/s} \]

### Part 2: Position at \( t = 3 \, \text{s} \)

1. **Horizontal Distance Covered:**
- The horizontal distance \( x \) can be calculated as:
\[ x = u_x \cdot t = 20 \cdot 3 = 60 \, \text{m} \]

2. **Vertical Distance Fallen:**
- The vertical distance \( y \) can be calculated using the equation of motion:
\[ y = u_y \cdot t + \frac{1}{2} g t^2 \]
\[ y = 0 + \frac{1}{2} \cdot 10 \cdot (3)^2 = \frac{1}{2} \cdot 10 \cdot 9 = 45 \, \text{m} \]

3. **Height from the Ground:**
- The height from the ground after falling is:
\[ \text{Height from ground} = \text{Initial height} - y = 50 - 45 = 5 \, \text{m} \]

4. **Position Coordinates:**
- The position coordinates at \( t = 3 \, \text{s} \) are:
\[ (x, y) = (60, 5) \]

### Part 3: Velocity Just Before Hitting the Ground

1. **Vertical Component Just Before Hitting the Ground:**
- Using energy conservation or kinematic equations:
\[ v_y^2 = u_y^2 + 2g h \]
\[ v_y^2 = 0 + 2 \cdot 10 \cdot 50 = 1000 \implies v_y = \sqrt{1000} = 10\sqrt{10} \, \text{m/s} \]

2. **Magnitude of the Overall Velocity Just Before Hitting the Ground:**
- The overall velocity \( v \) is:
\[ v = \sqrt{v_x^2 + v_y^2} = \sqrt{(20)^2 + (10\sqrt{10})^2} = \sqrt{400 + 1000} = \sqrt{1400} = 10\sqrt{14} \, \text{m/s} \]

### Part 4: Change in Velocity in 2 Seconds

1. **Change in Velocity:**
- The change in velocity \( \Delta v \) from \( t = 0 \) to \( t = 2 \, \text{s} \):
\[ \Delta v = v_{2s} - v_{0s} \]
- The velocity at \( t = 2 \, \text{s} \) is \( 20\sqrt{2} \, \text{m/s} \) and the initial velocity is \( 20 \, \text{m/s} \).
\[ \Delta v = 20\sqrt{2} - 20 \]

### Final Answers:
1. Velocity at \( t = 2 \, \text{s} \): \( 20\sqrt{2} \, \text{m/s} \)
2. Position at \( t = 3 \, \text{s} \): \( (60, 5) \)
3. Velocity just before hitting the ground: \( 10\sqrt{14} \, \text{m/s} \)
4. Change in velocity in 2 seconds: \( 20(\sqrt{2} - 1) \, \text{m/s} \)