A body is projected up such that its position vector varies with time as `r = { 3thati + (4 t - 5t^2)hatj}` m. Here, t is in seconds. Find the time and x-coordinate of particle when its y-coordinate is zero.

Explanation

To solve the problem, we need to find the time and x-coordinate of the particle when its y-coordinate is zero. Let's break it down step by step:

### Step 1: Identify the position vector
The position vector of the body is given as:
\[ \mathbf{r} = 3t \hat{i} + (4t - 5t^2) \hat{j} \]
Here, the x-coordinate is represented by the component along \(\hat{i}\) and the y-coordinate is represented by the component along \(\hat{j}\).

### Step 2: Extract the x and y coordinates
From the position vector, we can identify:
- \( x = 3t \)
- \( y = 4t - 5t^2 \)

### Step 3: Set the y-coordinate to zero
We need to find the time \( t \) when the y-coordinate is zero:
\[ 4t - 5t^2 = 0 \]

### Step 4: Factor the equation
We can factor out \( t \):
\[ t(4 - 5t) = 0 \]
This gives us two solutions:
1. \( t = 0 \)
2. \( 4 - 5t = 0 \) which simplifies to \( t = \frac{4}{5} \)

### Step 5: Solve for the x-coordinate
Now, we need to find the x-coordinate at the time when \( y = 0 \). We can use both values of \( t \) we found.

1. For \( t = 0 \):
\[ x = 3(0) = 0 \]

2. For \( t = \frac{4}{5} \):
\[ x = 3\left(\frac{4}{5}\right) = \frac{12}{5} \text{ m} \]

### Conclusion
The y-coordinate is zero at two instances: \( t = 0 \) seconds (initial position) and \( t = \frac{4}{5} \) seconds. The corresponding x-coordinates are:
- At \( t = 0 \): \( x = 0 \) m
- At \( t = \frac{4}{5} \): \( x = \frac{12}{5} \) m

### Final Answer
- Time when y-coordinate is zero: \( t = 0 \) seconds and \( t = \frac{4}{5} \) seconds
- Corresponding x-coordinates: \( x = 0 \) m and \( x = \frac{12}{5} \) m

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