A particle is projected from ground with velocity `20(sqrt2) m//s` at `45^@`. At what time particle is at height 15 m from ground? `(g = 10 m//s^2)`

Explanation

To solve the problem of finding the time at which a particle projected from the ground reaches a height of 15 meters, we can follow these steps:

### Step 1: Identify the given values
- Initial velocity \( u = 20\sqrt{2} \, \text{m/s} \)
- Angle of projection \( \theta = 45^\circ \)
- Height \( s_y = 15 \, \text{m} \)
- Acceleration due to gravity \( g = 10 \, \text{m/s}^2 \)

### Step 2: Calculate the vertical component of the initial velocity
The vertical component of the initial velocity \( u_y \) can be calculated using the formula:
\[ u_y = u \sin(\theta) \]
Substituting the values:
\[ u_y = 20\sqrt{2} \cdot \sin(45^\circ) = 20\sqrt{2} \cdot \frac{1}{\sqrt{2}} = 20 \, \text{m/s} \]

### Step 3: Use the kinematic equation for vertical motion
The equation for vertical motion is given by:
\[ s_y = u_y t + \frac{1}{2} a_y t^2 \]
Here, \( a_y = -g = -10 \, \text{m/s}^2 \) (since gravity acts downwards). Substituting the known values:
\[ 15 = 20t - \frac{1}{2} \cdot 10 t^2 \]
This simplifies to:
\[ 15 = 20t - 5t^2 \]

### Step 4: Rearrange the equation
Rearranging the equation gives:
\[ 5t^2 - 20t + 15 = 0 \]
Dividing the entire equation by 5:
\[ t^2 - 4t + 3 = 0 \]

### Step 5: Factor the quadratic equation
We can factor the quadratic equation:
\[ (t - 1)(t - 3) = 0 \]

### Step 6: Solve for \( t \)
Setting each factor to zero gives:
\[ t - 1 = 0 \quad \Rightarrow \quad t = 1 \, \text{s} \]
\[ t - 3 = 0 \quad \Rightarrow \quad t = 3 \, \text{s} \]

### Conclusion
The particle reaches a height of 15 meters at two different times: \( t = 1 \, \text{s} \) and \( t = 3 \, \text{s} \).

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