A lift, initially at rest on the ground, starts ascending with constant acceleration `8m//s^(2)` After 0.5 seconds, a bolt falls off the floor of the lift. The velocity of the lift at the instant the bolt hits the ground is m/s .[Take `g=10 m//s^(2) ]`

Explanation

To solve the problem step-by-step, let's break it down:

### Step 1: Understand the Given Information
- The lift starts from rest, so its initial velocity \( u = 0 \, \text{m/s} \).
- The acceleration of the lift \( a = 8 \, \text{m/s}^2 \).
- The time before the bolt falls off the lift \( t = 0.5 \, \text{s} \).
- The acceleration due to gravity \( g = 10 \, \text{m/s}^2 \).

### Step 2: Calculate the Velocity of the Lift When the Bolt Falls
To find the velocity of the lift at the moment the bolt falls off, we can use the formula for velocity under constant acceleration:
\[ v = u + at \]
Substituting the values:
\[ v = 0 + (8 \, \text{m/s}^2)(0.5 \, \text{s}) = 4 \, \text{m/s} \]
So, the velocity of the lift when the bolt falls off is \( 4 \, \text{m/s} \).

### Step 3: Calculate the Height of the Lift When the Bolt Falls
Next, we calculate the height of the lift at the moment the bolt falls off using the formula for distance traveled under constant acceleration:
\[ h = ut + \frac{1}{2} a t^2 \]
Substituting the values:
\[ h = 0 + \frac{1}{2} (8 \, \text{m/s}^2)(0.5 \, \text{s})^2 = \frac{1}{2} \cdot 8 \cdot 0.25 = 1 \, \text{m} \]
So, the height of the lift when the bolt falls off is \( 1 \, \text{m} \).

### Step 4: Determine the Time Taken for the Bolt to Hit the Ground
When the bolt falls, it will be in free fall. The initial velocity of the bolt at the moment it falls is equal to the velocity of the lift, which is \( 4 \, \text{m/s} \). We can use the following equation of motion to find the time \( t' \) it takes for the bolt to hit the ground:
\[ s = ut' + \frac{1}{2} g (t')^2 \]
Where \( s = 1 \, \text{m} \) (the height from which it falls), \( u = 4 \, \text{m/s} \), and \( g = 10 \, \text{m/s}^2 \). Rearranging gives:
\[ 1 = 4t' + \frac{1}{2} \cdot 10 \cdot (t')^2 \]
This simplifies to:
\[ 1 = 4t' + 5(t')^2 \]
Rearranging gives us a quadratic equation:
\[ 5(t')^2 + 4t' - 1 = 0 \]

### Step 5: Solve the Quadratic Equation
Using the quadratic formula \( t' = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
- Here, \( a = 5 \), \( b = 4 \), and \( c = -1 \).
\[ t' = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 5 \cdot (-1)}}{2 \cdot 5} \]
\[ t' = \frac{-4 \pm \sqrt{16 + 20}}{10} = \frac{-4 \pm \sqrt{36}}{10} = \frac{-4 \pm 6}{10} \]
Calculating the two possible values:
1. \( t' = \frac{2}{10} = 0.2 \, \text{s} \) (valid)
2. \( t' = \frac{-10}{10} = -1 \, \text{s} \) (not valid)

Thus, \( t' = 0.2 \, \text{s} \).

### Step 6: Calculate the Total Time for the Lift
The total time from the start until the bolt hits the ground is:
\[ t_{\text{total}} = 0.5 \, \text{s} + 0.2 \, \text{s} = 0.7 \, \text{s} \]

### Step 7: Calculate the Final Velocity of the Lift
Now, we can calculate the velocity of the lift at \( t = 0.7 \, \text{s} \):
\[ v_{\text{lift}} = u + at \]
Substituting the values:
\[ v_{\text{lift}} = 0 + (8 \, \text{m/s}^2)(0.7 \, \text{s}) = 5.6 \, \text{m/s} \]

### Final Answer
The velocity of the lift at the instant the bolt hits the ground is **5.6 m/s**.

---