How Can Theta Be Integrated into the Projectile Motion Equation?

AI Thread Summary
The discussion focuses on deriving an equation for the initial muzzle velocity of a projectile launcher, given various angles (theta) and recorded distances. The user has isolated the velocity components (Vix and Viy) using the equations for horizontal and vertical motion but struggles to integrate theta into the overall equation. They have calculated the time of flight and the x-component of velocity but are confused about how to incorporate the varying variables effectively. The conversation emphasizes the need for an additional equation to connect the unknowns of initial velocity and time, particularly in the y-direction. Overall, the integration of theta into the projectile motion equations remains a key challenge.
chillb0y
Messages
3
Reaction score
0
Projectile Motion Derivation

Homework Statement



I am supposed to derive an equation that ends up giving me a solution for the initial muzzle velocity of the projectile launcher. I am given different sets of theta and the recorded distance for the x distances. The height of the projectile launcher is also .25m.

Homework Equations



I used dx = Vixt, dy = viyt2 + (1/2)at2 to somewhat isolate the Vix and Viy variables.

The Attempt at a Solution



After isolating the variables, I plugged them into the a2 + b2 = c2 formula (cause the two components form a right angle to find the overall velocity)

i got this equation..

sq((dx2 + (dy-(1/2)at2)2)/t2) = Vi

the thing I am frustrated atm is that I am supposed to integrate theta into the equation but I cannot find a way to do this. it seems like the only varying variable would be dx, but now that i look at it again t should be a varying variable as well? and dy? so now it's more of a mess than an equation.

would love help, this has been giving me a headache for a while.

*edit: if it helps i calculated that a projectile i nthe projectile launcher is in there for only .224s.
 
Last edited:
Physics news on Phys.org
ok after some more analysis i figured out the velocity of the x component all the time (6.84m/s). now having ViX i need to derive a formula so i can use the time from the dy axis and plug it into the dx=vixt

somewhat like this

dy = viyt + .5at2
dx = vixt

Viy = tan theta vix (i am given vix and theta)
 
ok i think i got it

solve for Dx

dy = (tan theta vix)(dx/vix) - (g/2)(dx/vix)2
 
tell me projectile motion equationsin 3d
 
Hints:

1. The velocity in the x direction is constant. So if you know the time of flight, you can determine the distance as a function of the x component of initial velocity.

2. You have two unknowns in the above. They are initial velocity and time.

3. Come up with another equation that will have both unknowns present. To do this, work with the y directon.
 
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