My Math Forum projectile motion, how to get the equation for range in projectile motion

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 May 20th, 2014, 11:40 PM #1 Newbie   Joined: May 2014 From: australia Posts: 5 Thanks: 0 projectile motion, how to get the equation for range in projectile motion how do u derive or formulate, the equation for the trajectory in projectile motion. This is the equation how is this derived or created from the basics (like the displacement vector, velocity vector and vector constants; how is this equation created from them)?
May 21st, 2014, 05:38 PM   #2
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Quote:
 Originally Posted by sweer6 how do u derive or formulate, the equation for the trajectory in projectile motion. This is the equation how is this derived or created from the basics (like the displacement vector, velocity vector and vector constants; how is this equation created from them)?
This appears to be a range equation, not a projectile's path.

The simplest way to get a trajectory is to write:
$\displaystyle y = y_0 + v_{0y} t + \frac{1}{2} a_y t^2$

and
$\displaystyle x = x_0 + v_{0x} t + \frac{1}{2} a_x t^2$

Solve the x equation for t and sub it into the y equation. (Typically the acceleration in the x component will be zero and we can usually set x_0 to be 0 as well, so it isn't as bad as it looks.)

-Dan

 May 21st, 2014, 06:48 PM #3 Newbie   Joined: May 2014 From: australia Posts: 5 Thanks: 0 what i really need to find is the initial velocity, when i've been given the angle, the hight at launch (not 0) and the horizontal displacement at impact. So i need to use the above formula and solve for v, which is the initial velocity. But to do that i need to know how the above formula is derived or created? pls help Last edited by sweer6; May 21st, 2014 at 06:53 PM.
 May 21st, 2014, 07:09 PM #4 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,137 Thanks: 2381 Math Focus: Mainly analysis and algebra From the two equations you were given above: \begin{align*} y &= y_0 + v_{0y} t + \frac{1}{2} a_y t^2 \\ x &= x_0 + v_{0x} t + \frac{1}{2} a_x t^2 \\ \end{align*} The first thing to note is that $v_{0y}$ is the vertical component of the initial velocity, and $v_{0x}$ the horizontal component. Thus \begin{align*} v_{0y} &= v \cos \theta \\ v_{0x} &= v \sin \theta \\ \end{align*} Also, when the projectile lands $y = 0$ (I assume). This means that we can solve the first equation for $T$ the flight time. You can then put this value for $T$ into the second equation for $x$ to get the range of the flight. Thanks from topsquark
 May 21st, 2014, 11:00 PM #5 Newbie   Joined: May 2014 From: australia Posts: 5 Thanks: 0 Thanks but i still don't seem to get it, so i'll give you the question. The initial height is 1.95m (the intial horizontal displacement is 0), the angle of release is 35 degrees and the range of the projectile (or the horizontal displacement at impact) is 90.33m. How do i find the initial velocity?

 Tags equation, motion, motion-equation, projectile, range

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