A projectile is launched from a cliff of height H.  It is launched with an initial velocity v0 and angle 𝞱 above the horizontal.  Answer the following in terms of given quantities (in bold) and fundamental constants (“g”).  
What is its hangtime?


To determine the hangtime (total time of flight) of the projectile, we need to analyze the motion in the vertical direction. We'll apply kinematic equations to find the time when the projectile returns to the ground, taking into account both the initial vertical velocity and the height from which it was launched.

### Step 1: Break the motion into components
The initial velocity \(\mathbf{v_0}\) can be split into horizontal and vertical components:
- Vertical velocity: \(\mathbf{v_{0y}} = \mathbf{v_0} \sin \mathbf{\theta}\)
- Horizontal velocity: \(\mathbf{v_{0x}} = \mathbf{v_0} \cos \mathbf{\theta}\) (although this does not affect hangtime, it's useful to note for other calculations)

### Step 2: Use the kinematic equation for vertical motion
The general kinematic equation for vertical displacement is:

\[
y(t) = \mathbf{H} + \mathbf{v_{0y}} t - \frac{1}{2} g t^2
\]

Where:
- \(y(t) = 0\) when the projectile hits the ground (since the ground is at height 0),
- \(\mathbf{H}\) is the initial height of the cliff,
- \(\mathbf{v_{0y}} = \mathbf{v_0} \sin \mathbf{\theta}\),
- \(g\) is the acceleration due to gravity.

Setting \(y(t) = 0\), the equation becomes:

\[
0 = \mathbf{H} + \mathbf{v_0} \sin \mathbf{\theta} \cdot t - \frac{1}{2} g t^2
\]

This is a quadratic equation in terms of \(t\):

\[
\frac{1}{2} g t^2 - \mathbf{v_0} \sin \mathbf{\theta} \cdot t - \mathbf{H} = 0
\]

### Step 3: Solve the quadratic equation
We can use the quadratic formula to solve for \(t\), where \(a = \frac{1}{2} g\), \(b = -\mathbf{v_0} \sin \mathbf{\theta}\), and \(c = -\mathbf{H}\):

\[
t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

Substitute \(a\), \(b\), and \(c\) into the quadratic formula:

\[
t = \frac{-(-\mathbf{v_0} \sin \mathbf{\theta}) \pm \sqrt{(\mathbf{v_0} \sin \mathbf{\theta})^2 - 4 \cdot \frac{1}{2} g \cdot (-\mathbf{H})}}{2 \cdot \frac{1}{2} g}
\]

Simplifying:

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

### Step 4: Interpret the solution
Since time \(t\) must be positive, we take the positive root:

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

This is the **total hangtime** of the projectile.

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Would you like to explore more about how this equation is derived or applied? Let me know if you have any further questions!

Here are 5 related questions to deepen understanding:
1. How would the hangtime change if the cliff height is doubled?
2. How can we find the range of the projectile using the hangtime?
3. What would be the effect on hangtime if the angle of launch is increased?
4. Can the hangtime be determined if the initial velocity is horizontal?
5. How does air resistance alter the hangtime in real-world conditions?

**Tip:** In projectile motion problems, always separate horizontal and vertical components of velocity for easier analysis.

A projectile is launched from a cliff of height H. It is launched with an initial velocity v0 and angle 𝞱 above the horizontal. Answer the following in terms of given quantities (in bold) and fundamental constants (“g”). What is its hangtime?

Learn how to calculate the hangtime of a projectile launched from a cliff with an initial velocity v0 and angle 𝞱 above the horizontal. This solution applies kinematic equations and the quadratic formula to solve the problem, providing a clear breakdown. Suitable for advanced high school or college students.

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