Parabolic Equation assistance please.

  • Thread starter Thread starter GregD603
  • Start date Start date
  • Tags Tags
    Assistance
AI Thread Summary
The discussion revolves around understanding the Parabolic Equation used in projectile motion, specifically the equation y = tan(θ)x - (g / 2vo²cos²θ). The original poster, Greg, is struggling with applying this formula correctly and often ends up with complicated results. He seeks clarification on how the equation is derived from basic kinematic equations of motion. Another participant points out the need for a clearer connection between the kinematic equations and Greg's equation, suggesting that the derivation may not be straightforward. The conversation highlights the challenges in grasping the concepts behind projectile motion equations.
GregD603
Messages
2
Reaction score
0
I've been having difficulty understanding how the Parabolic Equation works... everytime I do a problem I get some really nasty looking numbers and often times it just doesn't work at all.
The equation I was taught is:

y = tan(θ)x - ( g / 2vo²cos²θ ) where vo = initial velocity vector

This equation was derived from: x = VxoT -or- T = x / Vxo (T = time) and
y = VyoT - ½gt²

Please help me understand how to apply this formula correctly.

Thank you very much,
Greg from Mass.
p.s. I know I didn't answer part three of the Template, however my question is more conceptual based rather than example. Pls don't delete
 
Physics news on Phys.org
Well for a given angle and initial velocity, you can find the vertical position at any horizontal position. I'm not too really sure what kind of answer you are looking for.
 
GregD603 said:
I've been having difficulty understanding how the Parabolic Equation works... everytime I do a problem I get some really nasty looking numbers and often times it just doesn't work at all.
The equation I was taught is:

y = tan(θ)x - ( g / 2vo²cos²θ ) where vo = initial velocity vector

This equation was derived from: x = VxoT -or- T = x / Vxo (T = time) and
y = VyoT - ½gt²

Please help me understand how to apply this formula correctly.

Thank you very much,
Greg from Mass.
p.s. I know I didn't answer part three of the Template, however my question is more conceptual based rather than example. Pls don't delete

Welcome to the PF, Greg. I'm not seeing where your top equation comes from. The bottom ones appear to be the kinematic equations of motion for constant acceleration (the "g" in this case), although they are a bit difficult to read without using LaTex:

y(t) = y_0 - \frac{1}{2} gt^2

How did you go from the kinematic equations to your first equation?
 
Thanks for the responses so quickly. And I"m sorry but I don't know what LaTex is, or where to obtain it. :(

My textbook tends to be super confusing about stuff that should be pretty simple, I guess.
The way it's described is:

t = x/Vxo is substituted into y = VyoT - ½gt² to obtain
y = Vyo(x/Vxo) - ½g(x/Vxo)² which can be rewritten as
y = (Vyo/Vxo)x - (g/2v2ocos2Θo)x2 or
y = (tanΘo)x - (g/2V2ocos2Θo)x2

and V2o is supposed to be initial velocity V squared.
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top