A projectile is launched from a cliff of height **H**.  It is launched with an initial velocity **v**0 and angle **𝞱** above the horizontal.  Answer the following in terms of given quantities (in bold) and fundamental constants (“g”).  

What is its hangtime?

What is its horizontal range?

When will it be descending at a 45 degree angle?

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**.

---

### 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**.

---

### 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**.

---

### 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:**
\[
t = \frac{v_0 (\sin(\theta) + \cos(\theta))}{g}
\]

---

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

A projectile is launched from a cliff of height **H**. It is launched with an initial velocity **v_0** and angle **𝞱** above the horizontal. Answer the following in terms of given quantities (in bold) and fundamental constants (“g”). What is its hangtime? What is its horizontal range? When will it be descending at a 45 degree angle?

Solve a projectile motion problem where a projectile is launched from a cliff of height H with an initial velocity v_0 and angle 𝞱. Learn how to calculate the hangtime, horizontal range, and the time when the projectile is descending at a 45-degree angle. This solution covers key physics concepts and kinematic formulas, suitable for advanced high school students.

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