Projectile Motion-acceleration/velocity

[TBE]

Hi everyone,
I am just starting to get into using physics and calculus, and have just been doing my own experiments to understand the concepts etc better.
I have formed an equation for displacement against time for a projectile and have differentiated it to get equation of velocity, then differentiated that to get equation of acceleration.
Equation: a=14.28t-17.56 (t=time in seconds)
I also know that at t=1.23, v=0; intitial v=10.80m/s; total displacement=8.86m at t=2.46; a due to g=9.81m/s/s.
I would just like to know how to work out the vertical velocity, as the decceleration is greater than 9.81 (I think).
I have spent the last few hours looking around the internet, but I'm not really sure what I'm looking for.
Thanks in advance.

 

[cupcuppycup]

please clarify all the given variables and what you are trying to find

 

[TBE]

Sorry,
I am trying to find the initial vertical velocity, ie the vertical velocity at time(t)=0.
equation for the acceleration of the projectile is a=14.28t+17.56
initial velocity=10.8m/s, ie v=10.8 @t=0; when v=0 (projectile at maximum height), t=1.23; displacement of projectile at t=1.23 is 4.43m.

I don't think I have explained this well, but basically I have an equation for acceleration that is a 'sloped' line, hence a is not constant. I want to know how I can find the initial vertical velocity. Because you would expect a due to g to be -9.81, I would normally multiply 1.23 by -9.81, but I can't see how that fits in with the equation that I've got.

I don't know if this helps...
Thanks in advance

 

[lackos]

im confused you say you are using projectile motion, however your equation for acceleration has a time dependant variable. this can't be true as the only acceleration on an object in projectile motion is that of gravity which is constant.

would you mind explaining how you got that equation? please

 

[TBE]

Hi lackos,
I got the equation by taking the second derivitive of the equaton I formed for the displacement over time of the projectile. I know it has a time dependent variable and this is what is confusing me.
I am beginning to think that I somehow didn't form the displacement over time equation properly. Would you expect the displacement over time equation of a projectile to be a cubic equation, rather than a quadratic?

Thanks

 

[ideasrule]

I am beginning to think that I somehow didn't form the displacement over time equation properly. Would you expect the displacement over time equation of a projectile to be a cubic equation, rather than a quadratic?

No. The displacement over time equation should be y=y0+v0*t-1/2*g*t^2. Differentiating that twice gives a=-g, which makes perfect sense for gravity.

 

[lackos]

okay I am not sure if this is answering your question, but i hope it does.

when you say "i formed" what do you mean, or better yet what did you do.

also the kinematic equations (which are used in projectile motion) are actually derived by integrating twice from a constant acceleration, with respect to time. this will only give you a quadratic equation for the displacement.

 

[PatrickRowe]

If you're doing what I think you're doing, you need to find your total initial velocity. In this case you state it as being 10.80m/s. In order to find the vertical component of the velocity you must use trigonometry. Assuming you are familiar with the basics, here, so correct me if I'm wrong, but imagine a triangle where the vector of your total velocity if the hypotenuse of a triangle. The vertical and horizontal components of this velocity can be found by using the same rules as you would any other triangle, to find the vertical then, sin(x) = opposite/hypotenuse, where x is the angle from the horizontal (ground level) at which you are "firing" your projectile. Opposite is the vertical velocity, and hypotenuse is your total velocity. From this we can say sin(x) = V(vertical)/V(total), rearranging gives us V(Vertical) = V(total)*Sin(x). This is your vertical velocity. The same logic can be applied to find the horizontal velocity component, however replacing sine for cosine, as found from Cosine(x) = Adjacent/Hypotenuse

 

[TBE]

Thanks everyone for your replies.
Ok, I fired a projectile that took 2.46s to go 8.86m. Therefore I plotted the points (0,0) [because at t=0, the displacement was 0m], (1.23,4.43) [because 1.23s is the turning point of the projectile ie v=0, and thus the displacement will be half the total displacement], (2.46,8.86) on a graph. I figured that that the equation should form a cubic, so I formed a cubic equation for displacement over time, then differentiated from there to get the equation for a.
I have not used any of the kinematic equations, so I'm thinking now that that is where I have gone wrong. I have just formed an equation like you would in maths. Looking at the kinematic equations now, I can see that they have been integrated from a constant acceleraton, so I will use them and see what I come up with.
The initial velocity, I just read from the graph when I differentiated it, so if my equations are wrong then that will also be.
Thanks for your help.