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# Average Velocity

## Displacement divided by time.

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Maximum Range

### Maximum Range

Source: http://www.flickr.com/photos/80682694@N05/7600910700

Created during the 12th century, the counterweight trebuchet has been used for hundreds of years to hurl objects long distance. Based on the conservation of energy and torque, trebuchets have launched objects weighing over 350lbs. To launch a projectile at furthest distance possible, only two things need to be taken into account; the difference in height between the launching point and the landing point of the projectile and the angle at which the projectile is released.

#### Why It Matters

• When projectile motion is discussed in physics it is usually under ideal conditions. Ideal conditions for projectile motion would be where there is no change in gravitational acceleration and air resistance is non-existent. Under these conditions the maximum range is defined by the following equation:

\begin{align*}Range = \frac{v \cos \theta}{g} \left(v \sin \theta + \sqrt{v^2 \sin^2 \theta + 2 gy_o}\right)\end{align*}

• In this equation, \begin{align*}\theta\end{align*} is the angle the projectile is launched at with respect to the horizontal, \begin{align*}v\end{align*} is the velocity and \begin{align*}y \end{align*} is the vertical distance the object is displaced through. Because the equation is assuming ideal conditions, slight adjustments need to be made for real life applications. For example, when launching artillery shells or in the Olympic shot put event, air resistance reduces the optimal angle that a projectile must be released at to achieve the optimal range. In these cases, the optimal range is slightly lower than 45°.
• Credit: SD Dirk
Source: http://www.flickr.com/photos/dirkhansen/6852350092/

Shot put is an example of projectile motion [Figure2]

#### Can You Apply It?

Using the information provided above, answer the following questions.

1. Using the following identity: \begin{align*}\sin \theta \cos \theta = \sin 2 \theta\end{align*}, show that the range equation given above reduces to \begin{align*}Range = \frac{v^2 \sin 2 \theta}{g}\end{align*} when the projectile lands at the same height it was launched?
2. Using the solution to the previous problem, if the initial velocity is doubled, the range is increased by a factor of?
3. The range equation for projectile motion helps validate that the motion of an object can be separated into motion along its coordinate axis, True or False?

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