Approach 1: Relative to the Elevator

1. Setup:
   • Bolt released at the top of the elevator.
   • Bolt’s acceleration relative to the elevator is \( g + a \).
   • Relative initial velocity of the bolt to the elevator is \( 0 \).
2. Equation: Use \( h = \frac{1}{2}(g + a)t^2 \).
3. Solve: \[ t = \sqrt{\frac{2h}{g + a}} \]

Approach 2: Relative to the Ground

1. Setup:
   • Elevator moves upward with acceleration \( a \): \[ y_e(t) = y_0 + ut + \frac{1}{2}a t^2 \]
   • Bolt falls downward with gravity: \[ y_b(t) = y_0 + h + ut – \frac{1}{2}g t^2 \]
2. Condition: Bolt hits the elevator floor when: \[ y_e(t) = y_b(t) \]
3. Equation: Substitute the expressions: \[ y_0 + ut + \frac{1}{2}a t^2 = y_0 + h + ut – \frac{1}{2}g t^2 \]
4. Solve: Simplify and solve for \( t \): \[ \frac{1}{2}(a + g)t^2 = h \quad \Rightarrow \quad t = \sqrt{\frac{2h}{g + a}} \]

Approach 3: Relative to the Gap

1. Setup:
   • Elevator moves upward a distance \( y_e = \frac{1}{2}a t^2 \).
   • Bolt falls downward a distance \( y_b = -\frac{1}{2}g t^2 \) (negative since downward).
   • Their total signed distances satisfy: \[ y_e – y_b = h \]
2. Equation: Substitute \( y_e = \frac{1}{2}a t^2 \) and \( y_b = -\frac{1}{2}g t^2 \): \[ \frac{1}{2}a t^2 – \left(-\frac{1}{2}g t^2\right) = h \]
3. Solve: Simplify and solve for \( t \): \[ \frac{1}{2}(a + g)t^2 = h \quad \Rightarrow \quad t = \sqrt{\frac{2h}{g + a}} \]