A lift accelerates downwards from rest at rate of `2 m//s^(2)`, starting `100 m` above the ground. After 3 sec, an object falls out of the lift. Which will reach the ground first? What is the time interval between their striking the ground?

Explanation

To solve the problem, we will break it down into steps:

### Step 1: Determine the position of the lift after 3 seconds
The lift accelerates downwards from rest at a rate of \(2 \, \text{m/s}^2\). We can use the equation of motion to find the distance covered by the lift in 3 seconds.

The equation of motion is given by:
\[ s = ut + \frac{1}{2} a t^2 \]
Where:
- \(s\) = distance covered
- \(u\) = initial velocity (which is \(0 \, \text{m/s}\) since it starts from rest)
- \(a\) = acceleration (\(2 \, \text{m/s}^2\))
- \(t\) = time (\(3 \, \text{s}\))

Substituting the values:
\[ s = 0 \cdot 3 + \frac{1}{2} \cdot 2 \cdot (3^2) = 0 + \frac{1}{2} \cdot 2 \cdot 9 = 9 \, \text{m} \]

### Step 2: Calculate the height of the lift above the ground after 3 seconds
Initially, the lift is \(100 \, \text{m}\) above the ground. After descending \(9 \, \text{m}\), the height of the lift above the ground is:
\[ \text{Height of lift} = 100 \, \text{m} - 9 \, \text{m} = 91 \, \text{m} \]

### Step 3: Determine the time taken for the object to reach the ground
When the object falls from the lift, it has an initial velocity equal to the velocity of the lift at that moment. The velocity of the lift after 3 seconds can be calculated using:
\[ v = u + at \]
Substituting the values:
\[ v = 0 + 2 \cdot 3 = 6 \, \text{m/s} \]

Now, the object falls from a height of \(91 \, \text{m}\) with an initial velocity of \(6 \, \text{m/s}\) and under the influence of gravity (\(g = 10 \, \text{m/s}^2\)). We can use the equation of motion:
\[ s = ut + \frac{1}{2} g t^2 \]
Where:
- \(s = 91 \, \text{m}\)
- \(u = 6 \, \text{m/s}\)
- \(g = 10 \, \text{m/s}^2\)

Rearranging gives:
\[ 91 = 6t + \frac{1}{2} \cdot 10 \cdot t^2 \]
This simplifies to:
\[ 91 = 6t + 5t^2 \]
Rearranging gives us a quadratic equation:
\[ 5t^2 + 6t - 91 = 0 \]

### Step 4: Solve the quadratic equation
Using the quadratic formula \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
- \(a = 5\)
- \(b = 6\)
- \(c = -91\)

Calculating the discriminant:
\[ D = b^2 - 4ac = 6^2 - 4 \cdot 5 \cdot (-91) = 36 + 1820 = 1856 \]

Now substituting into the quadratic formula:
\[ t = \frac{-6 \pm \sqrt{1856}}{2 \cdot 5} \]
Calculating \(\sqrt{1856} \approx 43.1\):
\[ t = \frac{-6 \pm 43.1}{10} \]
Taking the positive root:
\[ t = \frac{37.1}{10} \approx 3.71 \, \text{s} \]

### Step 5: Calculate the total time for the object
The total time taken by the object to reach the ground is:
\[ \text{Total time} = 3 \, \text{s} + 3.71 \, \text{s} = 6.71 \, \text{s} \]

### Step 6: Determine the time taken by the lift to reach the ground
Using the same equation of motion for the lift, which travels \(100 \, \text{m}\):
\[ 100 = 0 + \frac{1}{2} \cdot 2 \cdot t^2 \]
This simplifies to:
\[ 100 = t^2 \implies t = 10 \, \text{s} \]

### Step 7: Calculate the time interval between their striking the ground
The time interval between the object and the lift striking the ground is:
\[ \text{Time interval} = 10 \, \text{s} - 6.71 \, \text{s} = 3.29 \, \text{s} \]

### Final Answers
1. The object will reach the ground first.
2. The time interval between their striking the ground is approximately \(3.29 \, \text{s}\).