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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 over long distances. Using the principles of conservation of energy and torque, trebuchets have been engineered to launch objects weighing over 350 lbs. To launch a projectile at a target from the 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:

Range=vcosθg(vsinθ+v2sin2θ+2gyo)\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, θ\begin{align*}\theta\end{align*} is the angle at which the projectile is launched with respect to the horizontal, v\begin{align*}v\end{align*} is the velocity of the projectile, and y\begin{align*}y\end{align*} is the vertical distance through which the object is displaced. Because the equation assumes ideal conditions, slight adjustments need to be made for real life applications. For example, when considering launching artillery shells or throwing shot puts in the Olympic event, air resistance reduces the optimal angle that a projectile must be released at to achieve the optimal range. In these cases, the optimal angle 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: sinθcosθ=sin2θ\begin{align*}\sin \theta \cos \theta = \sin 2 \theta\end{align*}, show that the range equation given above reduces to Range=v2sin2θg\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, by what factor does the range increase if the velocity is doubled?
3. True or False: The range equation for projectile motion helps validate that the motion of an object can be separated into motion along both of its coordinate axis.

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