
New Users Post up here and introduce yourself! 
 LinkBack  Thread Tools  Display Modes 
May 20th, 2014, 10:40 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, 04:38 PM  #2  
Math Team Joined: May 2013 From: The Astral plane Posts: 1,559 Thanks: 601 Math Focus: Wibbly wobbly timeywimey stuff.  Quote:
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.) This page is a good start to help you with your general problem. Dan  
May 21st, 2014, 05:48 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 05:53 PM. 
May 21st, 2014, 06:09 PM  #4 
Math Team Joined: Dec 2013 From: Colombia Posts: 6,782 Thanks: 2197 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. 
May 21st, 2014, 10:00 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, motionequation, projectile, range 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Differential equation that involve projectile motion ;w;  Haejang  Differential Equations  7  May 6th, 2014 07:48 AM 
projectile motion  shalini maniarasan  Calculus  1  April 25th, 2014 01:26 AM 
Projectile Motion Problem  edwinandrew  Physics  1  February 26th, 2014 07:27 PM 
Physics: Projectile Motion/Relative Motion  edwinandrew  Physics  0  February 26th, 2014 04:30 PM 
Projectile Motion  symmetry  Algebra  1  June 19th, 2007 10:26 PM 