We can solve this problem using the kinematic equations of motion. Here are the steps:

1. Set up the problem:

• Initial height: $h_0 = 100 \, \text{m}$
• Initial velocity: $v_0 = 75 \, \text{m/s}$
• Angle of projection: $\theta = 32^\circ$ (below the horizontal)
• Acceleration due to gravity: $g = 9.8 \, \text{m/s}^2$

2. Break the initial velocity into components:

• Horizontal velocity ($v_{0x}$): $v_{0x} = v_0 \cdot \cos(\theta) = 75 \cdot \cos(32^\circ)$
• Vertical velocity ($v_{0y}$) (downward since it's below the horizontal): $v_{0y} = v_0 \cdot \sin(\theta) = 75 \cdot \sin(32^\circ)$

3. Apply the kinematic equation for vertical motion:

The vertical motion can be modeled using the following equation: $y = y_0 + v_{0y} \cdot t + \frac{1}{2} g t^2$ Where:

• $y = 0 \, \text{m}$ (since the projectile hits the ground),
• $y_0 = 100 \, \text{m}$,
• $v_{0y} = 75 \cdot \sin(32^\circ) \, \text{m/s}$,
• $g = 9.8 \, \text{m/s}^2$.

Rearranging this equation, we get: $0 = 100 + (75 \cdot \sin(32^\circ)) \cdot t + \frac{1}{2} (-9.8) t^2$ This is a quadratic equation in $t$, and we can solve it to find the time of flight. Let's calculate the values and solve for $t$.The projectile takes approximately 10.13 seconds to hit the ground.

Would you like more details on this, or do you have any further questions?

Here are five related questions:

1. How far horizontally does the projectile travel before hitting the ground?
2. What is the speed of the projectile just before it hits the ground?
3. How would the time of flight change if the projectile were shot at a steeper angle?
4. What if the initial velocity were increased—how would that affect the time of flight?
5. How would air resistance affect the time to hit the ground?

Tip: Always break the velocity into horizontal and vertical components when solving projectile problems.