A rocket is fired upward from the earth's surface such that it creates an acceleration of `20 m//s^(2)`. If after 5 s its engine is switched off, the maximum height of the rochet from the earth's surface would be

Options

A. 250 m
B. 500 m
C. 750 m
D. 1000 m

Answer

To solve the problem step by step, we will break it down into two main parts: the time when the rocket is accelerating and the time after the engine is switched off.

### Step 1: Calculate the velocity of the rocket after 5 seconds of acceleration
The rocket is fired with an upward acceleration of \(20 \, \text{m/s}^2\) for \(5\) seconds. We can use the formula for velocity under constant acceleration:

\[ v = u + at \]

Where:
- \(v\) = final velocity
- \(u\) = initial velocity (which is \(0 \, \text{m/s}\) since it starts from rest)
- \(a\) = acceleration (\(20 \, \text{m/s}^2\))
- \(t\) = time (\(5 \, \text{s}\))

Substituting the values:

\[ v = 0 + (20 \, \text{m/s}^2)(5 \, \text{s}) = 100 \, \text{m/s} \]

### Step 2: Calculate the height gained during the first 5 seconds
We can use the formula for distance traveled under constant acceleration:

\[ s = ut + \frac{1}{2} a t^2 \]

Where:
- \(s\) = distance traveled
- \(u\) = initial velocity (\(0 \, \text{m/s}\))
- \(a\) = acceleration (\(20 \, \text{m/s}^2\))
- \(t\) = time (\(5 \, \text{s}\))

Substituting the values:

\[ s = (0)(5) + \frac{1}{2} (20)(5^2) = 0 + \frac{1}{2} (20)(25) = 250 \, \text{m} \]

### Step 3: Calculate the additional height gained after the engine is switched off
After \(5\) seconds, the engine is switched off, and the rocket will continue to rise until it reaches its maximum height. At this point, the initial velocity for this phase is \(100 \, \text{m/s}\) (the velocity we calculated in Step 1), and the acceleration due to gravity will act downward at \(-9.8 \, \text{m/s}^2\).

Using the formula for maximum height when the final velocity is \(0\):

\[ v^2 = u^2 + 2as \]

Where:
- \(v\) = final velocity (\(0 \, \text{m/s}\))
- \(u\) = initial velocity (\(100 \, \text{m/s}\))
- \(a\) = acceleration (\(-9.8 \, \text{m/s}^2\))
- \(s\) = additional height gained (let's call it \(y\))

Rearranging the formula to solve for \(s\):

\[ 0 = (100)^2 + 2(-9.8)y \]

This simplifies to:

\[ 0 = 10000 - 19.6y \]

Solving for \(y\):

\[ 19.6y = 10000 \implies y = \frac{10000}{19.6} \approx 510.2 \, \text{m} \]

### Step 4: Calculate the total maximum height
Now, we can find the total height \(H\) by adding the height gained during the first \(5\) seconds and the additional height gained after the engine is switched off:

\[ H = s + y = 250 \, \text{m} + 510.2 \, \text{m} \approx 760.2 \, \text{m} \]

### Final Answer
The maximum height of the rocket from the Earth's surface would be approximately \(760.2 \, \text{m}\).

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