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Boundless Physics
Two-Dimensional Kinematics
Projectile Motion
Physics Textbooks Boundless Physics Two-Dimensional Kinematics Projectile Motion
Physics Textbooks Boundless Physics Two-Dimensional Kinematics
Physics Textbooks Boundless Physics
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Physics
Concept Version 10
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General Launch Angle

The initial launch angle (0-90 degrees) of an object in projectile motion dictates the range, height, and time of flight of that object.

Learning Objective

  • Choose the appropriate equation to find range, maximum height, and time of flight


Key Points

    • If the same object is launched at the same initial velocity, the height and time of flight will increase proportionally to the initial launch angle.
    • An object launched into projectile motion will have an initial launch angle anywhere from 0 to 90 degrees.
    • The range of an object, given the initial launch angle and initial velocity is found with: $R=\frac{v_i^2 \sin2\theta_i}{g}$.
    • The maximum height of an object, given the initial launch angle and initial velocity is found with:$h=\frac{v_i^2\sin^2\theta_i}{2g}$ .
    • The time of flight of an object, given the initial launch angle and initial velocity is found with: $T=\frac{2v_i\sin\theta}{g}$.
    • The angle of reach is the angle the object must be launched at in order to achieve a specific distance: $\theta=\frac12\sin^{-1}(\frac{gd}{v^2})$.

Term

  • trajectory

    The path of a body as it travels through space.


Examples

    • You have an object that is being launched into projectile motion. The initial velocity of the object is: $v_i=10$ m/s Find the range, maximum height and time of flight for the object at the following initial launch angles: a. 15° b. 45° c. 70° We will work out part (a) here, but you will have to do parts (b) and (c) on your own. The answers for those parts can be found at the bottom of the example:
    • Range: $\displaystyle{R=\frac{v_i^2\sin2\theta_i}{g}\\ R=\frac{10(\frac{m}{s})^2\sin2*{15^{\circ}}}{9.81(\frac{m}{s})}= \frac{10(\frac{m}{s})^2\sin{30^{\circ}}}{9.81(\frac{m}{s})}\\ R=\frac{10(\frac{m}{s})^2*0.5}{9.81(\frac{m}{s})}= 5.1m}$
    • Maximum Height: 𝗁=𝗏𝟤𝗂sin𝟤θ𝗂𝟤𝗀𝗁=𝟣𝟢(𝗆𝗌)𝟤sin𝟤𝟣𝟧∘𝟤∗𝟫.𝟪𝟣(𝗆𝗌)=𝟣𝟢(𝗆𝗌)𝟤sin𝟤𝟣𝟧∘𝟣𝟫.𝟨𝟤(𝗆𝗌)𝗁=𝟣𝟢(𝗆𝗌)𝟤∗𝟢.𝟢𝟨𝟩𝟣𝟫.𝟨𝟤(𝗆𝗌)=𝟢.𝟥𝟣𝗆\small{\sf{h=\frac{v_i^2\sin^2\theta_i}{2g}}}\\ \small{\sf{h=\frac{10(\frac{m}{s})^2\sin^2{15^{\circ}}}{2*9.81(\frac{m}{s})}}}= \small{\sf{\frac{10(\frac{m}{s})^2\sin^2{15^{\circ}}}{19.62(\frac{m}{s})}}}\\ \small{\sf{h=\frac{10(\frac{m}{s})^2*0.067}{19.62(\frac{m}{s})}}}= \small{\sf{0.31m}}
    • Time of Flight:$\small{\sf{T=\frac{2v_i\sin\theta_i}{g}}}\\ \small{\sf{T=\frac{2*10(\frac{m}{s})\sin{15^{\circ}}}{9.81(\frac{m}{s})}}}= \small{\sf{\frac{20(\frac{m}{s})*0.5}{9.81(\frac{m}{s})}}}\\ \small{\sf{T=\frac{10(\frac{m}{s})}{9.81(\frac{m}{s})}}}= \small{\sf{1.01s}}$b. 45 °Range - 10.1 mHeight - 2.54 mTime of Flight - 1.44 sc. 70 °Range - 6.55 mHeight - 4.50 m Time of Flight - 1.91 s

Full Text

Projectile motion is a form of motion where an object moves in a bilaterally symmetrical, parabolic path. The path that the object follows is called its trajectory. Projectile motion only occurs when there is one force applied at the beginning of the trajectory, after which the only interference is from gravity.

One of the key components of projectile motion and the trajectory that it follows is the initial launch angle. This angle can be anywhere from 0 to 90 degrees. The angle at which the object is launched dictates the range, height, and time of flight it will experience while in projectile motion. shows different paths for the same object launched at the same initial velocity at different launch angles. As you can see from the figure, the larger the initial launch angle, the closer the object comes to maximum height and the longer the flight time. The largest range will be experienced at a launch angle up to 45 degrees.

Launch Angle

The launch angle determines the range and maximum height that an object will experience after being launched. This image shows that path of the same object being launched at the same velocity but different angles.

The range, maximum height, and time of flight can be found if you know the initial launch angle and velocity, using the following equations:

$\small{\sf{R=\frac{v_i^2\sin2\theta_i}{g}}}\\ \small{\sf{h=\frac{v_i^2\sin^2\theta_i}{2g}}}\\ \small{\sf{T=\frac{2v_isin\theta}{g}}}$

Where R - Range, h - maximum height, T - time of flight, vi - initial velocity, θi - initial launch angle, g - gravity.

Now that we understand how the launch angle plays a major role in many other components of the trajectory of an object in projectile motion, we can apply that knowledge to making an object land where we want it. If there is a certain distance, d, that you want your object to go and you know the initial velocity at which it will be launched, the initial launch angle required to get it that distance is called the angle of reach. It can be found using the following equation:

$\small{\sf{\theta=\frac12sin^{-1}(\frac{gd}{v^2})}}$

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