To find the maximum height \( H \) and the total time of flight \( T \) for a particle projected with an initial speed \( v_0 \) at an angle \( \theta \) with the horizontal, we can use the following kinematic equations:
### Maximum Height \( H \)
The maximum height is reached when the vertical component of the velocity becomes zero. The initial vertical velocity is \( v_{0y} = v_0 \sin \theta \). The acceleration in the vertical direction is \( -g \) (acceleration due to gravity).
Using the kinematic equation:
\[ v_y^2 = v_{0y}^2 + 2a_y(y - y_0) \]
At maximum height, \( v_y = 0 \), and \( y - y_0 = H \):
\[ 0 = (v_0 \sin \theta)^2 + 2(-g)H \]
\[ H = \frac{(v_0 \sin \theta)^2}{2g} \]
### Total Time of Flight \( T \)
The total time of flight is the time taken for the particle to return to the ground. This can be found by considering the vertical motion and using the initial vertical velocity and the acceleration due to gravity.
The vertical displacement after time \( T \) is zero (returns to ground level):
\[ y = y_0 + v_{0y} t + \frac{1}{2} a_y t^2 \]
Since the displacement \( y - y_0 = 0 \):
\[ 0 = v_0 \sin \theta \cdot T - \frac{1}{2} g T^2 \]
\[ 0 = T \left( v_0 \sin \theta - \frac{1}{2} g T \right) \]
Solving for \( T \):
\[ T = \frac{2 v_0 \sin \theta}{g} \]
### Summary
- **Maximum Height \( H \):**
\[ H = \frac{(v_0 \sin \theta)^2}{2g} \]
- **Total Time of Flight \( T \):**
\[ T = \frac{2 v_0 \sin \theta}{g} \]
