A particle is projected from a tower of height 25 m with velocity `20sqrt(2)m//s` at `45^@`. (a) Find the time when particle strikes with ground. (b) The horizontal distance from the foot of tower where it strikes. (c) Also find the velocity at the time of collision.

Explanation

To solve the problem step by step, we will break it down into three parts as stated in the question:

### Given Data:
- Height of the tower (h) = 25 m
- Initial velocity (u) = \(20\sqrt{2}\) m/s
- Angle of projection (θ) = 45°
- Acceleration due to gravity (g) = 9.8 m/s² (acting downwards)

### Step 1: Find the time when the particle strikes the ground.

1. **Resolve the initial velocity into horizontal and vertical components:**
- \(u_x = u \cdot \cos(θ) = 20\sqrt{2} \cdot \cos(45°) = 20\sqrt{2} \cdot \frac{1}{\sqrt{2}} = 20 \, \text{m/s}\)
- \(u_y = u \cdot \sin(θ) = 20\sqrt{2} \cdot \sin(45°) = 20\sqrt{2} \cdot \frac{1}{\sqrt{2}} = 20 \, \text{m/s}\)

2. **Use the second equation of motion for vertical displacement:**
The vertical displacement \(y\) is given by:
\[ y = u_y t - \frac{1}{2} g t^2 \]
Here, \(y = -25\) m (downward displacement), \(u_y = 20\) m/s, and \(g = 9.8\) m/s².
\[ -25 = 20t - \frac{1}{2} \cdot 9.8 t^2 \]
Rearranging gives:
\[ 4.9t^2 - 20t - 25 = 0 \]

3. **Solve the quadratic equation using the quadratic formula:**
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \(a = 4.9\), \(b = -20\), and \(c = -25\).
\[ t = \frac{20 \pm \sqrt{(-20)^2 - 4 \cdot 4.9 \cdot (-25)}}{2 \cdot 4.9} \]
\[ t = \frac{20 \pm \sqrt{400 + 490}}{9.8} \]
\[ t = \frac{20 \pm \sqrt{890}}{9.8} \]
\[ t = \frac{20 \pm 29.83}{9.8} \]
Taking the positive root:
\[ t = \frac{49.83}{9.8} \approx 5.08 \, \text{s} \]

### Step 2: Find the horizontal distance from the foot of the tower where it strikes.

1. **Calculate the horizontal distance (d):**
\[ d = u_x \cdot t \]
\[ d = 20 \cdot 5.08 \approx 101.6 \, \text{m} \]

### Step 3: Find the velocity at the time of collision.

1. **Calculate the vertical component of the velocity just before impact (v_y):**
Using the equation:
\[ v_y^2 = u_y^2 + 2g h \]
\[ v_y^2 = 20^2 + 2 \cdot 9.8 \cdot 25 \]
\[ v_y^2 = 400 + 490 = 890 \]
\[ v_y = \sqrt{890} \approx 29.83 \, \text{m/s} \]

2. **The horizontal component of the velocity remains constant:**
\[ v_x = 20 \, \text{m/s} \]

3. **Calculate the resultant velocity (v):**
\[ v = \sqrt{v_x^2 + v_y^2} \]
\[ v = \sqrt{20^2 + 29.83^2} = \sqrt{400 + 890} = \sqrt{1290} \approx 35.91 \, \text{m/s} \]

4. **Find the direction of the velocity:**
\[ \tan(\theta) = \frac{v_y}{v_x} = \frac{29.83}{20} \]
\[ \theta = \tan^{-1}\left(\frac{29.83}{20}\right) \approx 33.84^\circ \]

### Final Answers:
(a) Time when the particle strikes the ground: **5.08 s**
(b) Horizontal distance from the foot of the tower: **101.6 m**
(c) Velocity at the time of collision: **35.91 m/s at an angle of 33.84° to the horizontal.**