A ball is projected from the top of a tower with a velocity `3hati+4hatj+5hatk ms^(-1)`,where `hati+hatj+hatk` are unit vectors along east, north and vertical upwards respectively. If the height of the tower is `30 m`, its time is `(g=10ms^(-1))`

Options

A. 5
B. 3
C. 0.3
D. 0.5

Explanation

To solve the problem of finding the time it takes for a ball projected from the top of a tower to reach the ground, we can break down the motion into its vertical and horizontal components.

### Step-by-Step Solution:

1. **Identify the Given Information:**
- Initial velocity of the ball: \( \vec{u} = 3 \hat{i} + 4 \hat{j} + 5 \hat{k} \, \text{m/s} \)
- Height of the tower: \( h = 30 \, \text{m} \)
- Acceleration due to gravity: \( g = 10 \, \text{m/s}^2 \)

2. **Focus on the Vertical Motion:**
- The vertical component of the initial velocity is \( u_z = 5 \, \text{m/s} \) (upward).
- The downward acceleration due to gravity is \( a = -g = -10 \, \text{m/s}^2 \).

3. **Use the Equation of Motion:**
- The equation of motion that relates displacement, initial velocity, time, and acceleration is:
\[ s = ut + \frac{1}{2} a t^2 \]
- Here, \( s = -30 \, \text{m} \) (since the displacement is downward), \( u = 5 \, \text{m/s} \), and \( a = -10 \, \text{m/s}^2 \).

4. **Substituting the Values:**
- Substitute the known values into the equation:
\[ -30 = 5t - \frac{1}{2} \cdot 10 \cdot t^2 \]
- Simplifying this gives:
\[ -30 = 5t - 5t^2 \]
- Rearranging the equation:
\[ 5t^2 - 5t - 30 = 0 \]
- Dividing the entire equation by 5:
\[ t^2 - t - 6 = 0 \]

5. **Factoring the Quadratic Equation:**
- We can factor the quadratic equation:
\[ (t - 3)(t + 2) = 0 \]
- This gives us two possible solutions for \( t \):
\[ t = 3 \quad \text{or} \quad t = -2 \]

6. **Selecting the Valid Solution:**
- Since time cannot be negative, we discard \( t = -2 \).
- Therefore, the time taken for the ball to reach the ground is:
\[ t = 3 \, \text{seconds} \]

### Final Answer:
The time taken for the ball to reach the ground is **3 seconds**.