Assertion At height 20 m from ground , velocity of a projectile is `v = (20 hati + 10 hatj) ms^(-1)`. Here, `hati` is horizontal and `hatj` is vertical. Then, the particle is at the same height after 4s.
Reason Maximum height of particle from ground is 40m (take, `g = 10 ms^(-2))`

Options

A. If both Asseration and Reason are correct and Reason is the correct explanation of Assertion
B. If both Assertion and Reason are correct but Reason is not the correct explanation of Assertion
C. If Assertion is true but Reason is false
D. If Assertion is false but Reason is true

Explanation

To solve the problem step by step, let's analyze the assertion and reason provided in the question.

### Step 1: Understanding the Given Information
The velocity of the projectile at a height of 20 m is given as:
\[ \vec{v} = 20 \hat{i} + 10 \hat{j} \, \text{m/s} \]
This indicates that the horizontal component of the velocity (\(v_x\)) is 20 m/s and the vertical component of the velocity (\(v_y\)) is 10 m/s.

### Step 2: Analyzing Vertical Motion
The vertical motion of the projectile is influenced by gravity. The vertical velocity will change over time due to the acceleration caused by gravity (\(g = 10 \, \text{m/s}^2\)).

Using the equation of motion:
\[ v_y = u_y - gt \]
Where:
- \(v_y\) is the final vertical velocity,
- \(u_y\) is the initial vertical velocity (10 m/s),
- \(g\) is the acceleration due to gravity (10 m/s²),
- \(t\) is the time.

### Step 3: Finding Time to Reach Maximum Height
At maximum height, the vertical component of the velocity becomes zero:
\[ 0 = 10 - 10t \]
Solving for \(t\):
\[ 10t = 10 \]
\[ t = 1 \, \text{s} \]

### Step 4: Time to Return to the Same Height
The projectile takes 1 second to reach the maximum height from 20 m and will take another 1 second to return back to 20 m. Therefore, the total time to go up and come back down to the same height (20 m) is:
\[ t_{\text{total}} = 1 + 1 = 2 \, \text{s} \]

### Step 5: Evaluating the Assertion
The assertion states that the particle is at the same height after 4 seconds. Since we calculated that it takes 2 seconds to return to the height of 20 m, the assertion is **incorrect**.

### Step 6: Finding Maximum Height
To find the maximum height reached by the projectile, we can use the equation:
\[ v^2 = u^2 - 2gs \]
Where:
- \(v\) is the final vertical velocity (0 m/s at maximum height),
- \(u\) is the initial vertical velocity (10 m/s),
- \(g\) is the acceleration due to gravity (10 m/s²),
- \(s\) is the height gained from the point of 20 m.

Substituting the values:
\[ 0 = (10)^2 - 2(10)s \]
\[ 0 = 100 - 20s \]
\[ 20s = 100 \]
\[ s = 5 \, \text{m} \]

### Step 7: Total Maximum Height
The maximum height from the ground is:
\[ \text{Maximum height} = 20 \, \text{m} + 5 \, \text{m} = 25 \, \text{m} \]

### Step 8: Evaluating the Reason
The reason states that the maximum height of the particle from the ground is 40 m. Since we calculated it to be 25 m, the reason is also **incorrect**.

### Conclusion
Both the assertion and the reason are incorrect.

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