A rocket is fired vertically up from the ground with a resultant vertical acceleration of `10 m//s^2.` The fuel is finished in 1 min and it continues to move up. (a) What is the maximum height reached? (b) Afte2r how much time from then will the maximum height be reached?(Take `g= 10 m//s^2`)

Explanation

To solve the problem step by step, we will break it down into two parts: (a) calculating the maximum height reached by the rocket, and (b) determining the time taken to reach that maximum height after the fuel is finished.

### Part (a): Maximum Height Reached

1. **Determine the initial conditions**:
- The rocket is fired with an initial velocity \( u = 0 \) m/s.
- The resultant vertical acceleration \( a = 10 \) m/s².
- The fuel burns for \( t = 1 \) minute = 60 seconds.

2. **Calculate the final velocity when the fuel runs out**:
We can use the equation of motion:
\[ v = u + at \]
Substituting the known values:
\[ v = 0 + (10 \, \text{m/s}^2)(60 \, \text{s}) = 600 \, \text{m/s} \]
So, the final velocity \( v \) when the fuel runs out is \( 600 \, \text{m/s} \).

3. **Calculate the height reached during the fuel burn (H1)**:
We can use the equation:
\[ s = ut + \frac{1}{2}at^2 \]
Substituting the values:
\[ H_1 = (0)(60) + \frac{1}{2}(10)(60^2) = 0 + \frac{1}{2}(10)(3600) = 18000 \, \text{m} \]
Thus, \( H_1 = 18000 \, \text{m} = 18 \, \text{km} \).

4. **Calculate the additional height (H2) after the fuel runs out**:
After the fuel runs out, the rocket will continue to rise until its velocity becomes zero. The only force acting on it is gravity, which decelerates it at \( g = 10 \, \text{m/s}^2 \).

Using the formula:
\[ H_2 = \frac{v^2}{2g} \]
Substituting the values:
\[ H_2 = \frac{(600)^2}{2(10)} = \frac{360000}{20} = 18000 \, \text{m} \]
Thus, \( H_2 = 18000 \, \text{m} = 18 \, \text{km} \).

5. **Calculate the total maximum height**:
The total maximum height \( H \) is the sum of \( H_1 \) and \( H_2 \):
\[ H = H_1 + H_2 = 18000 + 18000 = 36000 \, \text{m} = 36 \, \text{km} \]

### Part (b): Time to Reach Maximum Height After Fuel Runs Out

1. **Determine the time taken to reach maximum height after fuel runs out**:
We know the initial velocity at the point where the fuel runs out is \( u = 600 \, \text{m/s} \), the final velocity \( v = 0 \, \text{m/s} \), and the acceleration \( a = -10 \, \text{m/s}^2 \) (since gravity acts downwards).

Using the equation:
\[ v = u + at \]
Rearranging for \( t \):
\[ t = \frac{v - u}{a} = \frac{0 - 600}{-10} = \frac{-600}{-10} = 60 \, \text{s} \]

### Final Answers
(a) The maximum height reached is \( 36 \, \text{km} \).
(b) The time taken to reach the maximum height after the fuel runs out is \( 60 \, \text{s} \).