A rocket is fired vertically from the ground. It moves upwards with a constant acceleration `10m//s^(2)` after 30 sec the fuel is finished. After what time from the instant of firing the rocket will attain the maximum height? `g=10m//s^(2)`:-

Explanation

To solve the problem step by step, we will break it down into parts.

### Step 1: Calculate the distance traveled by the rocket while the fuel is burning.

The rocket accelerates upwards with a constant acceleration \( a = 10 \, \text{m/s}^2 \) for \( t = 30 \, \text{s} \). The initial velocity \( u = 0 \) since it starts from rest.

Using the equation of motion:
\[ s = ut + \frac{1}{2} a t^2 \]
Substituting the values:
\[ s = 0 \cdot 30 + \frac{1}{2} \cdot 10 \cdot (30)^2 \]
\[ s = \frac{1}{2} \cdot 10 \cdot 900 = 5 \cdot 900 = 4500 \, \text{m} \]

### Step 2: Calculate the velocity of the rocket at the end of the fuel burn.

Using the equation:
\[ v = u + at \]
Substituting the values:
\[ v = 0 + 10 \cdot 30 = 300 \, \text{m/s} \]

### Step 3: Determine the time taken to reach the maximum height after the fuel is finished.

Once the fuel is finished, the rocket will continue to move upwards but will decelerate due to gravity. The acceleration due to gravity \( g = 10 \, \text{m/s}^2 \) acts downwards.

Using the equation:
\[ v = u + at \]
Here, at maximum height, the final velocity \( v = 0 \), the initial velocity \( u = 300 \, \text{m/s} \), and the acceleration \( a = -10 \, \text{m/s}^2 \) (negative because it is acting downwards).

Setting up the equation:
\[ 0 = 300 - 10t \]
Solving for \( t \):
\[ 10t = 300 \]
\[ t = 30 \, \text{s} \]

### Step 4: Calculate the total time from the instant of firing to the maximum height.

The total time taken to reach maximum height is the sum of the time during which the fuel was burning and the time taken to reach the maximum height after the fuel is consumed:
\[ \text{Total time} = 30 \, \text{s} + 30 \, \text{s} = 60 \, \text{s} \]

### Final Answer:
The rocket will attain its maximum height after **60 seconds** from the instant of firing.

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