A particle is projected with a velocity 10 m//s at an angle `37^(@)` to the horizontal. Find the location at which the particle is at a height 1 m from point of projection.

Explanation

To solve the problem of finding the location of a particle projected with a velocity of 10 m/s at an angle of 37 degrees to the horizontal when it reaches a height of 1 meter, we can follow these steps:

### Step 1: Identify the components of the initial velocity
The initial velocity \( u \) can be resolved into horizontal and vertical components:
- \( u_x = u \cos \theta \)
- \( u_y = u \sin \theta \)

Given:
- \( u = 10 \, \text{m/s} \)
- \( \theta = 37^\circ \)

Using trigonometric values:
- \( \cos 37^\circ = \frac{4}{5} \)
- \( \sin 37^\circ = \frac{3}{5} \)

Calculating the components:
\[ u_x = 10 \cdot \frac{4}{5} = 8 \, \text{m/s} \]
\[ u_y = 10 \cdot \frac{3}{5} = 6 \, \text{m/s} \]

### Step 2: Write the equation for vertical motion
The vertical position \( y \) of the particle as a function of time \( t \) is given by:
\[ y = u_y t - \frac{1}{2} g t^2 \]
where \( g \) is the acceleration due to gravity (approximately \( 10 \, \text{m/s}^2 \)).

### Step 3: Set the height to 1 meter
We need to find the time when the particle reaches a height of 1 meter:
\[ 1 = 6t - \frac{1}{2} \cdot 10 t^2 \]
This simplifies to:
\[ 1 = 6t - 5t^2 \]
Rearranging gives:
\[ 5t^2 - 6t + 1 = 0 \]

### Step 4: Solve the quadratic equation
Using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 5 \), \( b = -6 \), and \( c = 1 \).
\[ t = \frac{6 \pm \sqrt{(-6)^2 - 4 \cdot 5 \cdot 1}}{2 \cdot 5} \]
\[ t = \frac{6 \pm \sqrt{36 - 20}}{10} \]
\[ t = \frac{6 \pm \sqrt{16}}{10} \]
\[ t = \frac{6 \pm 4}{10} \]
Calculating the two possible values for \( t \):
1. \( t = \frac{10}{10} = 1 \, \text{s} \)
2. \( t = \frac{2}{10} = 0.2 \, \text{s} \)

### Step 5: Find the horizontal distance at these times
The horizontal distance \( x \) is given by:
\[ x = u_x t \]
Calculating for both times:
1. For \( t = 1 \, \text{s} \):
\[ x = 8 \cdot 1 = 8 \, \text{m} \]
2. For \( t = 0.2 \, \text{s} \):
\[ x = 8 \cdot 0.2 = 1.6 \, \text{m} \]

### Final Result
The particle reaches a height of 1 meter at two locations:
- At \( x = 8 \, \text{m} \) (on the way up)
- At \( x = 1.6 \, \text{m} \) (on the way down)