To solve the problem of finding the time of flight of a particle projected under gravity to reach a height of 9.81 m, we can follow these steps:
### Step 1: Identify the Given Values
- Initial velocity, \( U = 29.43 \, \text{m/s} \)
- Angle of projection, \( \theta = 30^\circ \)
- Acceleration due to gravity, \( g = 9.81 \, \text{m/s}^2 \)
- Height to be reached, \( s = 9.81 \, \text{m} \)
### Step 2: Calculate the Vertical Component of the Initial Velocity
The vertical component of the initial velocity \( U_y \) can be calculated using:
\[
U_y = U \sin(\theta)
\]
Substituting the values:
\[
U_y = 29.43 \sin(30^\circ) = 29.43 \times 0.5 = 14.715 \, \text{m/s}
\]
### Step 3: Use the Equation of Motion
We can use the equation of motion for vertical displacement:
\[
s = U_y t + \frac{1}{2} (-g) t^2
\]
Here, \( s = 9.81 \, \text{m} \), \( U_y = 14.715 \, \text{m/s} \), and \( g = 9.81 \, \text{m/s}^2 \). The equation becomes:
\[
9.81 = 14.715 t - \frac{1}{2} (9.81) t^2
\]
This simplifies to:
\[
9.81 = 14.715 t - 4.905 t^2
\]
### Step 4: Rearranging the Equation
Rearranging the equation gives us:
\[
4.905 t^2 - 14.715 t + 9.81 = 0
\]
### Step 5: Solve the Quadratic Equation
Using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
- Here, \( a = 4.905 \), \( b = -14.715 \), and \( c = 9.81 \).
Calculating the discriminant:
\[
D = b^2 - 4ac = (-14.715)^2 - 4 \times 4.905 \times 9.81
\]
Calculating \( D \):
\[
D = 216.56 - 192.12 = 24.44
\]
Now substituting back into the quadratic formula:
\[
t = \frac{14.715 \pm \sqrt{24.44}}{2 \times 4.905}
\]
Calculating \( \sqrt{24.44} \):
\[
\sqrt{24.44} \approx 4.944
\]
Thus:
\[
t = \frac{14.715 \pm 4.944}{9.81}
\]
Calculating the two possible values of \( t \):
1. \( t_1 = \frac{14.715 + 4.944}{9.81} \approx 2.0 \, \text{s} \)
2. \( t_2 = \frac{14.715 - 4.944}{9.81} \approx 1.0 \, \text{s} \)
### Step 6: Conclusion
The particle reaches the height of 9.81 m at two different times: \( t_1 \approx 2.0 \, \text{s} \) and \( t_2 \approx 1.0 \, \text{s} \).
