A bullet is projected upwards from the top of a tower of height 90 m with the velocity `30 ms^(-1)` making an angle `30^(@)` with the horizontal. Find the time taken by it to reach the ground is `(g=10 ms^(-2))`

Answer

To solve the problem of a bullet projected upwards from the top of a tower, we will follow these steps:

### Step 1: Identify the given values
- Height of the tower (h) = 90 m
- Initial velocity (u) = 30 m/s
- Angle of projection (θ) = 30°
- Acceleration due to gravity (g) = 10 m/s²

### Step 2: Resolve the initial velocity into its components
The initial velocity can be broken down into horizontal and vertical components:
- Vertical component (u_y) = u * sin(θ) = 30 * sin(30°) = 30 * 0.5 = 15 m/s
- Horizontal component (u_x) = u * cos(θ) = 30 * cos(30°) = 30 * (√3/2) = 15√3 m/s (not needed for this calculation)

### Step 3: Set up the equation for vertical motion
Since the bullet is projected upwards and then falls down, we will use the equation of motion for vertical displacement:
\[ s_y = u_y t - \frac{1}{2} g t^2 \]
Where:
- \( s_y \) = vertical displacement (which will be -90 m since it falls down)
- \( u_y \) = initial vertical velocity (15 m/s)
- \( g \) = acceleration due to gravity (10 m/s²)
- \( t \) = time in seconds

### Step 4: Substitute the values into the equation
Substituting the known values into the equation:
\[ -90 = 15t - \frac{1}{2} \cdot 10 \cdot t^2 \]
This simplifies to:
\[ -90 = 15t - 5t^2 \]

### Step 5: Rearrange the equation
Rearranging gives us:
\[ 5t^2 - 15t - 90 = 0 \]
Dividing the entire equation by 5:
\[ t^2 - 3t - 18 = 0 \]

### Step 6: Factor the quadratic equation
We can factor the equation:
\[ (t - 6)(t + 3) = 0 \]

### Step 7: Solve for t
Setting each factor to zero gives:
1. \( t - 6 = 0 \) → \( t = 6 \) seconds
2. \( t + 3 = 0 \) → \( t = -3 \) seconds (not valid since time cannot be negative)

### Conclusion:
The time taken by the bullet to reach the ground is **6 seconds**.

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