A bullet is fired vertically up from a 400 m tall tower with a speed 80 m/s. If g is taken as `10 m//s^(2),` the time taken by the bullet to reach the ground will be

Options

A. 8 s
B. 16 s
C. 20 s
D. 24 s

Explanation

To solve the problem of a bullet fired vertically from a 400 m tall tower with an initial speed of 80 m/s, we need to calculate the total time taken by the bullet to reach the ground. We can break this down into several steps:

### Step 1: Analyze the motion of the bullet
The bullet is fired upwards with an initial velocity (u) of 80 m/s. It will first ascend until it reaches its maximum height, then it will descend back down to the ground.

### Step 2: Calculate the time taken to reach the maximum height (T1)
At the maximum height, the final velocity (v) will be 0 m/s. We can use the first equation of motion:
\[ v = u + at \]
Where:
- \( v = 0 \) m/s (final velocity at maximum height)
- \( u = 80 \) m/s (initial velocity)
- \( a = -g = -10 \) m/s² (acceleration due to gravity, acting downwards)

Substituting the values:
\[ 0 = 80 - 10T_1 \]
\[ 10T_1 = 80 \]
\[ T_1 = \frac{80}{10} = 8 \text{ seconds} \]

### Step 3: Calculate the maximum height reached (H)
We can use the second equation of motion to find the maximum height (H):
\[ H = uT_1 + \frac{1}{2} a T_1^2 \]
Substituting the values:
\[ H = 80 \times 8 + \frac{1}{2} \times (-10) \times (8^2) \]
\[ H = 640 - 320 = 320 \text{ meters} \]

### Step 4: Calculate the total height from which the bullet falls
The total height from which the bullet falls to the ground is the height of the tower plus the maximum height reached:
\[ \text{Total height} = 400 \text{ m (tower height)} + 320 \text{ m (maximum height)} = 720 \text{ m} \]

### Step 5: Calculate the time taken to fall to the ground (T3)
Now, we will calculate the time taken to fall from this height using the second equation of motion:
\[ S = uT_3 + \frac{1}{2} a T_3^2 \]
Where:
- \( S = 720 \) m (total height)
- \( u = 0 \) m/s (initial velocity when falling)
- \( a = 10 \) m/s² (acceleration due to gravity)

Substituting the values:
\[ 720 = 0 \times T_3 + \frac{1}{2} \times 10 \times T_3^2 \]
\[ 720 = 5T_3^2 \]
\[ T_3^2 = \frac{720}{5} = 144 \]
\[ T_3 = \sqrt{144} = 12 \text{ seconds} \]

### Step 6: Calculate the total time taken (T_total)
The total time taken by the bullet to reach the ground is the sum of the time taken to reach the maximum height (T1), the time taken to descend back to the tower height (T2, which is equal to T1), and the time taken to fall to the ground (T3):
\[ T_{\text{total}} = T_1 + T_1 + T_3 = 8 + 8 + 12 = 28 \text{ seconds} \]

### Final Answer
The total time taken by the bullet to reach the ground is **28 seconds**.

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