# Calculating initial velocity using kinematic equations

1. May 17, 2015

### cmonstre

Hi! I'm working on a problem for class and am having trouble figuring out how to calculate the initial velocity, which I need in order to calculate acceleration and from there the value of an unknown force. I thought I was on the right track, but my professor reminded me that I need to use the instantaneous initial velocity, not the average velocity. The problem is below:

1. The problem statement, all variables and given/known data

A 0.15kg object is launched from the ground and moves under the influence of gravity as well as a second force (wind). The wind force is constant, but magnitude and direction are unknown. We are given a table of values (points) from the plot of the trajectory and are asked to calculate the force of the wind on the ball.
The table looks like this:
t(s), x(m), y(m)
t=0, x=0, y=0
t=.5, x=3.46, y=4.59
t=1, x=6.72, y=6.85
t=1.5, x=9.81, y=6.79
t=2, x=12.70, y=4.40

The graph is obviously a parabola.

2. Relevant equations
I'll put the kinematic equations down for reference:
deltax= 1/2 (Vf - V0)t
deltax = Vft - 1/2at2
deltax= V0t + 1/2at2
Vf - V0 = at
Vf2= V02 + 2a(deltax)

3. The attempt at a solution
Alright. I know that the values in the table can give me information to calculate displacement, instantaneous velocities, and time, from which I can calculate acceleration using the kinematic equations. Since I also know the mass of the object, I can then use the F=ma equation to calculate the force of the wind.

I know that if the force on an object is constant, the acceleration must be constant (because the mass is obviously constant)…so the acceleration of the object wouldn't change across the trajectory. I think this means that the vertical acceleration (Ay) would be the acceleration due to gravity PLUS the acceleration of the wind, and the horizontal acceleration (Ax) would just be the acceleration of the wind. Unless of course the wind acceleration only affects the horizontal acceleration, and then the vertical acceleration is just acceleration due to gravity.

My original thought process was to calculate initial horizontal velocity, and then calculate acceleration from there. The issue I ran into is that using the values from the table only gave me average velocity across that interval. Then I thought that I could solve for the angle of launch for a better approximation of V0.
I used the x value of 6.72m and the y value of 6.85m, then used tan(theta)=6.85/6.72 to solve for the launch angle, which gave me approximately 45.5 degrees. The initial horizontal velocity would then be V0x= V0*cos(45.5), and the initial vertical velocity would be V0y=V0*sin(45.5). My professor stopped me here and reminded me that this is still an approximation though and told me to construct a system of equations that would allow me to solve for the exact initial velocity. I'm not sure where to start with the system of equations, because I only have the displacement and time variables as I would run into the same issue of having an average velocity if I tried to calculate Vf.

Once I have the value for initial velocity, however, I would be able to solve for acceleration using
deltax= V0t + 1/2at2
and then multiply the acceleration by the mass of the projectile to get Fx. From there it would just be solving for the unknown force.

Thanks for the help! I know this is ultimately pretty basic, but this is my first physics class!

2. May 18, 2015

### BvU

Hello Cm, welcome to PF !

Pretty clear posting, but you had me puzzled with "The graph is obviously a parabola". Was that part of the problem formulation, or did you add it ?
And which graph are you referring to (I am so primitive that I first think of 2-D graphs, so x(t), y(t) or y(x) ) ?
(and when I draw them, they all look like parabolas quite well -- also x(t) ! In fact the first two look like perfect parabolas...)

This may be your first physics class, but your prof is really asking some tough questions here ! Are you allowed to use some modern tools, such as fitting a parabola through the points using the trendline gadget in Excel ? If not, then you have to do some real work to get out reasonable answers. Fortunately, the time steps are all 0.5 seec, so that helps.

is not applicable: The vertical acceleration is clearly different from -4.9 m/s2.

sounds good, but unfortunately the motions in the x-direction and in the y-direction are completely independent in this exercise.

You also make a strange choice when you pick t=1 points to calculate the launch angle ! After all, that angle changes with time fairly rapidly (check with the angle at t=0.5 s!)

(That is probably what triggered prof to intervene. You have a good one there, so if I were you I would listen carefully !)

--

To get you going: my hint is that x(t) and y(t) look like nearly perfect parabolas. As you know, a parabola is completely determined by three parameters.
In SUVAT equations: y(t) = y0 + v0 t + 0.5 a t2.
So three points should be enough to construct them (just like two points yield a line -- two parameters).
You have five points, so you can do a little averaging -- easy with a calculator .
x(t) is a good candidate to begin with.

Let us know how you are getting on.