To solve the problem, we need to find the horizontal range of a body projected from the top of a tower with a given initial velocity. Let's break down the solution step by step.
### Step 1: Identify the components of the initial velocity
The initial velocity \( \mathbf{u} \) is given as:
\[
\mathbf{u} = 3 \hat{i} + 4 \hat{j} + 5 \hat{k} \, \text{m/s}
\]
Here, \( \hat{i} \) is the east direction, \( \hat{j} \) is the north direction, and \( \hat{k} \) is the vertical direction (upwards).
### Step 2: Determine the height of the tower
The height of the tower is given as \( h = 30 \, \text{m} \).
### Step 3: Use the vertical motion equation to find the time of flight
We will use the equation of motion in the vertical direction:
\[
s = ut + \frac{1}{2} a t^2
\]
Where:
- \( s = -30 \, \text{m} \) (the displacement is downward)
- \( u = 5 \, \text{m/s} \) (initial vertical velocity)
- \( a = -10 \, \text{m/s}^2 \) (acceleration due to gravity)
Substituting the values into the equation:
\[
-30 = 5t - \frac{1}{2} \cdot 10 t^2
\]
This simplifies to:
\[
-30 = 5t - 5t^2
\]
Rearranging gives:
\[
5t^2 - 5t - 30 = 0
\]
Dividing through by 5:
\[
t^2 - t - 6 = 0
\]
### Step 4: Factor the quadratic equation
Factoring the quadratic:
\[
(t - 3)(t + 2) = 0
\]
Thus, the solutions for \( t \) are:
\[
t = 3 \, \text{s} \quad \text{or} \quad t = -2 \, \text{s}
\]
Since time cannot be negative, we have:
\[
t = 3 \, \text{s}
\]
### Step 5: Calculate the horizontal velocity
The horizontal components of the initial velocity are:
\[
u_x = 3 \, \text{m/s} \quad \text{(east direction)}
\]
\[
u_y = 4 \, \text{m/s} \quad \text{(north direction)}
\]
The resultant horizontal velocity \( u_h \) can be calculated using the Pythagorean theorem:
\[
u_h = \sqrt{u_x^2 + u_y^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \, \text{m/s}
\]
### Step 6: Calculate the horizontal range
The horizontal range \( R \) can be calculated using:
\[
R = u_h \cdot t
\]
Substituting the values:
\[
R = 5 \, \text{m/s} \cdot 3 \, \text{s} = 15 \, \text{m}
\]
### Final Answer
The horizontal range of the body on the ground is:
\[
\boxed{15 \, \text{m}}
\]
