To solve this projectile motion problem, we need to calculate three things:

1. The maximum height reached by the projectile.
2. The total time the projectile spends in the air.
3. The horizontal distance the projectile travels (range).

Let's break the problem into parts and use the following equations of motion.

Step 1: Find the maximum height

The vertical component of the initial velocity is: $v_{0y} = v_0 \cdot \sin(\theta)$ where:

• $v_0 = 80 \, \text{m/s}$ (initial speed),
• $\theta = 35^\circ$ (angle of projection),
• $\sin(35^\circ)$ is the sine of 35 degrees.

Now, the maximum height $H$ is found using the formula: $H = \frac{{v_{0y}^2}}{2g}$ where $g = 9.8 \, \text{m/s}^2$ is the acceleration due to gravity.

Step 2: Calculate the total time of flight

The time of flight $T$ is given by the formula: $T = \frac{2 \cdot v_{0y}}{g}$ This equation gives the total time the projectile spends in the air.

Step 3: Find the horizontal distance (range)

The horizontal component of the initial velocity is: $v_{0x} = v_0 \cdot \cos(\theta)$ The horizontal distance (range) is: $R = v_{0x} \cdot T$ where $T$ is the total time of flight calculated in Step 2.

Let me compute these values.The results for the projectile are as follows:

1. Maximum height: $107.43 \, \text{m}$
2. Total time of flight: $9.36 \, \text{seconds}$
3. Horizontal distance (range): $613.68 \, \text{m}$

Do you want more details or have any questions?

Here are 5 related questions to expand your understanding:

1. How would the maximum height change if the angle of projection increased to 45 degrees?
2. What happens to the total time of flight if the initial speed is reduced by half?
3. How would air resistance affect the range of the projectile?
4. What is the relationship between the launch angle and the range of a projectile?
5. How does increasing the gravitational acceleration impact the motion of the projectile?

Tip: For maximum range, a projectile is typically launched at an angle of 45 degrees in the absence of air resistance.