An elevator is accelerating upward at a rate of `6ft/sec^(2)` when a bolt from its celling falls to the floor of the lift (Distance=9.5feet). The time taken (in seconds) by the falling bolt to hit the floor is (take `g=32ft//sec^(2)`)

Explanation

To solve the problem of a bolt falling from the ceiling of an upward-accelerating elevator, we can follow these steps:

### Step 1: Identify the effective acceleration
The elevator is accelerating upward at a rate of \(6 \, \text{ft/s}^2\) and the acceleration due to gravity \(g\) is \(32 \, \text{ft/s}^2\). Since the bolt is falling downwards while the elevator is accelerating upwards, we can find the effective acceleration acting on the bolt.

\[ g_{\text{effective}} = g - a = 32 \, \text{ft/s}^2 - 6 \, \text{ft/s}^2 = 26 \, \text{ft/s}^2 \]

### Step 2: Use the equation of motion
We can use the second equation of motion to find the time taken for the bolt to hit the floor. The equation is:

\[ s = ut + \frac{1}{2} g_{\text{effective}} t^2 \]

Where:
- \(s\) is the distance fallen (9.5 feet),
- \(u\) is the initial velocity (0 ft/s, since the bolt starts from rest),
- \(g_{\text{effective}}\) is the effective acceleration (26 ft/s²),
- \(t\) is the time in seconds.

Substituting the known values into the equation:

\[ 9.5 = 0 \cdot t + \frac{1}{2} \cdot 26 \cdot t^2 \]

This simplifies to:

\[ 9.5 = 13 t^2 \]

### Step 3: Solve for \(t^2\)
Now, we can rearrange the equation to solve for \(t^2\):

\[ t^2 = \frac{9.5}{13} \]

### Step 4: Calculate \(t\)
Now, we can calculate \(t\):

\[ t^2 = \frac{9.5}{13} \approx 0.730769 \]

Taking the square root:

\[ t = \sqrt{0.730769} \approx 0.854 \, \text{seconds} \]

### Final Answer
The time taken by the falling bolt to hit the floor is approximately \(0.854\) seconds.

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