Let's break down each part of the problem step by step.

1. Hangtime (Total time of flight)

The motion of the projectile can be split into two components: vertical and horizontal. To find the hangtime, we focus on the vertical motion.

The vertical position at any time $t$ is given by:

$y(t) = H + v_0 \sin(\theta) t - \frac{1}{2} g t^2$

The projectile hits the ground when $y(t) = 0$. Setting this equation to zero and solving for $t$:

$H + v_0 \sin(\theta) t - \frac{1}{2} g t^2 = 0$

This is a quadratic equation in $t$:

$\frac{1}{2} g t^2 - v_0 \sin(\theta) t - H = 0$

Solving for $t$ using the quadratic formula:

$t = \frac{-(-v_0 \sin(\theta)) \pm \sqrt{(-v_0 \sin(\theta))^2 - 4 \times \frac{1}{2} g \times (-H)}}{2 \times \frac{1}{2} g}$

Simplifying:

$t = \frac{v_0 \sin(\theta) \pm \sqrt{(v_0 \sin(\theta))^2 + 2 g H}}{g}$

Since time must be positive, we choose the positive root:

$t = \frac{v_0 \sin(\theta) + \sqrt{(v_0 \sin(\theta))^2 + 2 g H}}{g}$

This is the total hangtime.

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2. Horizontal Range

The horizontal range $R$ is the horizontal distance the projectile covers during its entire flight.

The horizontal position at any time $t$ is given by:

$x(t) = v_0 \cos(\theta) t$

The total range $R$ occurs when $t =$ hangtime (calculated in the previous part). Therefore:

$R = v_0 \cos(\theta) \times \frac{v_0 \sin(\theta) + \sqrt{(v_0 \sin(\theta))^2 + 2 g H}}{g}$

This is the horizontal range.

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3. When will it be descending at a 45-degree angle?

At a 45-degree angle of descent, the vertical and horizontal components of the velocity will be equal in magnitude but opposite in direction (since it's descending).

The vertical velocity at any time $t$ is:

$v_y(t) = v_0 \sin(\theta) - g t$

The horizontal velocity remains constant throughout the flight:

$v_x = v_0 \cos(\theta)$

At a 45-degree angle of descent, $v_y = -v_x$. Therefore:

$v_0 \sin(\theta) - g t = -v_0 \cos(\theta)$

Solving for $t$:

$v_0 \sin(\theta) + v_0 \cos(\theta) = g t$

$t = \frac{v_0 (\sin(\theta) + \cos(\theta))}{g}$

This gives the time when the projectile is descending at a 45-degree angle.

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Summary of Answers:

1. Hangtime: $t = \frac{v_0 \sin(\theta) + \sqrt{(v_0 \sin(\theta))^2 + 2 g H}}{g}$

2. Horizontal Range: $R = v_0 \cos(\theta) \times \frac{v_0 \sin(\theta) + \sqrt{(v_0 \sin(\theta))^2 + 2 g H}}{g}$

3. Time when descending at 45 degrees: $t = \frac{v_0 (\sin(\theta) + \cos(\theta))}{g}$

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Would you like more details on any part of this solution or any clarifications? Here are five questions you can explore further:

1. How does the height of the cliff affect the hangtime?
2. What would the horizontal range be if the launch angle was zero?
3. How