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, we need to determine the velocity of the lift at the instant the bolt hits the ground after falling from the lift. Here’s a step-by-step breakdown of the solution:

### Step 1: Determine the height of the lift after 0.5 seconds
The lift starts from rest and accelerates upwards with a constant acceleration of \(8 \, \text{m/s}^2\). We can use the equation of motion to find the distance traveled by the lift in 0.5 seconds.

Using the formula:
\[ s = ut + \frac{1}{2} a t^2 \]
Where:
- \(s\) = distance traveled by the lift
- \(u\) = initial velocity = \(0 \, \text{m/s}\) (since it starts from rest)
- \(a\) = acceleration = \(8 \, \text{m/s}^2\)
- \(t\) = time = \(0.5 \, \text{s}\)

Substituting the values:
\[ s = 0 \times 0.5 + \frac{1}{2} \times 8 \times (0.5)^2 \]
\[ s = 0 + \frac{1}{2} \times 8 \times 0.25 \]
\[ s = 0 + 1 = 1 \, \text{m} \]
So, the lift has ascended \(1 \, \text{m}\) after \(0.5\) seconds.

### Step 2: Determine the velocity of the lift at \(t = 0.5\) seconds
We can find the velocity of the lift at \(t = 0.5\) seconds using the formula:
\[ v = u + at \]
Where:
- \(v\) = final velocity
- \(u\) = initial velocity = \(0 \, \text{m/s}\)
- \(a\) = acceleration = \(8 \, \text{m/s}^2\)
- \(t\) = time = \(0.5 \, \text{s}\)

Substituting the values:
\[ v = 0 + 8 \times 0.5 \]
\[ v = 4 \, \text{m/s} \]
Thus, the velocity of the lift at \(t = 0.5\) seconds is \(4 \, \text{m/s}\).

### Step 3: Determine the time taken by the bolt to hit the ground
When the bolt falls from the lift, it has an initial upward velocity of \(4 \, \text{m/s}\) (the velocity of the lift at that moment). The bolt will fall a distance of \(1 \, \text{m}\) (the height of the lift).

Using the equation of motion for the bolt:
\[ s = ut + \frac{1}{2} (-g) t^2 \]
Where:
- \(s = -1 \, \text{m}\) (downward displacement)
- \(u = 4 \, \text{m/s}\) (initial upward velocity)
- \(g = 10 \, \text{m/s}^2\) (acceleration due to gravity)

Substituting the values:
\[ -1 = 4t - \frac{1}{2} \times 10 t^2 \]
\[ -1 = 4t - 5t^2 \]
Rearranging gives us:
\[ 5t^2 - 4t - 1 = 0 \]

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

Thus, the time taken by the bolt to hit the ground is \(1 \, \text{s}\).

### Step 5: Calculate the total time from the start until the bolt hits the ground
The total time from the start until the bolt hits the ground is:
\[ t_{\text{total}} = 0.5 \, \text{s} + 1 \, \text{s} = 1.5 \, \text{s} \]

### Step 6: Determine the final velocity of the lift at \(t = 1.5\) seconds
Now we can find the final velocity of the lift using the formula:
\[ v = u + at \]
Where:
- \(u = 0 \, \text{m/s}\) (initial velocity)
- \(a = 8 \, \text{m/s}^2\)
- \(t = 1.5 \, \text{s}\)

Substituting the values:
\[ v = 0 + 8 \times 1.5 \]
\[ v = 12 \, \text{m/s} \]

### Final Answer
The velocity of the lift at the instant the bolt hits the ground is \(12 \, \text{m/s}\).

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