How does substituting t in the y-coordinate equation affect the outcome?

  • Thread starter Thread starter bentley4
  • Start date Start date
AI Thread Summary
Substituting t in the y-coordinate equation using the x-coordinate equation simplifies the problem and leads to the correct answer. The discussion highlights confusion around why substituting v or both v and t does not yield the same results. It emphasizes that t must be eliminated because it is a variable, while v is a constant. The maximum range is achieved at a projection angle of 45 degrees, as derived from the range formula. Ultimately, understanding the need to eliminate t clarifies how to determine the maximum value of x.
bentley4
Messages
66
Reaction score
0
attachment.php?attachmentid=43340&stc=1&d=1328027012.jpg


In this solution the t in the y-coordinate equation is substituted using the x-coordinate equation and ultimately leads to the answer.

My questions:
1. Why don't I get the same answer when I substitute the v or v and t instead?
2. How am I supposed to know to substitute t in this example and not v?
 

Attachments

  • probl14.jpg
    probl14.jpg
    32.1 KB · Views: 1,293
Physics news on Phys.org


I don't see how you can get it by keeping the t. You don't know t, so how can you tell the maximum value of x that way? I get
x = gt²/(2*tanΘ)
and can't tell what combination of t and Θ provide the maximum x.
It would be good to see your calc.

There is no need to eliminate the v; it is a constant. But you must eliminate the variable t.
 


Well, the problem is quite easy to approach.
Range is given by u^2 sin(2a) / g, where a is the projection angle.
Since -1<sin a<1, the max. value for a sine function = 1. This occurs when the angle is 90 degrees or .5pi radians.
So, for a fixed u:
2a = 90
a = 45 degrees.
 


Delphi51 said:
I don't see how you can get it by keeping the t. You don't know t, so how can you tell the maximum value of x that way? I get
x = gt²/(2*tanΘ)
and can't tell what combination of t and Θ provide the maximum x.
It would be good to see your calc.

There is no need to eliminate the v; it is a constant. But you must eliminate the variable t.

Ok, now I understand. Thnx!
 
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}...
I was thinking using 2 purple mattress samples, and taping them together, I do want other ideas though, the main guidelines are; Must have a volume LESS than 1600 cubic centimeters, and CAN'T exceed 25 cm in ANY direction. Must be LESS than 1 kg. NO parachutes. NO glue or Tape can touch the egg. MUST be able to take egg out in less than 1 minute. Grade A large eggs will be used.
Back
Top