# Problem in projectile motion

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1. Dec 30, 2015

### CherryWine

PROBLEM REGARDING PROJECTILE MOTION
1.

I have to find the implicit equation for minimum and maximum initial velocity needed in order for water to end up in the reservoir. I have tried solving it by firstly expressing flight time from the equation for range of projectile (time in terms of D and Vx). I obtain t=6D/(v0*cosx45) - for minimum velocity. Then I substituted this in the equation of y(t) and obtained the trajectory equation. From that equation y(x), I expressed initial velocity and obtained V0=sqrt((D^2*g)/(2(cosx)^2*deltaY-2(cosx)^2*tanx*D)), and from there I obtained Vmin=2,44*sqrt(D*g) and Vmax=2,65*sqrt(D*g) while the correct answers are Vmin=3*sqrt(D*g) and Vmax=3,13*sqrt(D*g). I couldn't figure what was wrong in my method.

Edit: Please note that in the trajectory equation D means the range (6D and 7D), I did not use just one D for calculation. Also in trigonometric functions x is the angle=45 degrees. DeltaY is 2D.

2. Relevant equations
D=V0*cosx*t
t=D/(V0*cosx)
deltaY=V0*sinx*t+1/2*gt^2

2. Dec 30, 2015

### Fightfish

Sorry, I can't quite follow your train of thought. Could you could show us the equation of the trajectory, $y(x)$, that you found?
Note that the requirement that the projectile lands in the water sets the following 2 constraints: (1) $y(6D) \geq 2D$ and (2) $y(7D) < 2D$. Imposing these constraints will lead to two inequalities that will get you the minimum and maximum $v_{0}$ allowed.

3. Dec 30, 2015

### CherryWine

The trajectory equation I obtained is y(x)=v0*sinθ*((6D)/(v0*cosθ))+1/2*g*((6D^2*g)/(2v0^2*(cosθ)^2)). Which then simplifies to y(x)=tanθ*6D+((6D^2*g^2)/(2*v0^2*(costheta)^2)).

4. Dec 30, 2015

### Fightfish

Given that $\theta = 45^{o}$, you could have made your life a lot easier. What you have is not a trajectory equation - you have already evaluated the y-position at that particular point, but that works too - if your equation was correct. Remember that $g$ acts downwards (and it shouldn't be squared in the expression as well?)

5. Dec 30, 2015

### CherryWine

I don't think that you can use the inequality y(6D)>=2D because that could mean that at x=6D, y can be larger than 2D, and then the possibility of the projectile overshooting the reservoir stays open. I think that using strict equations y(6D)=2D and y(7D)=2D would yield the correct answer.

Sorry for the squared g, but the trajectory equation is not the main thing here, it is the equation for Vo that I obtained from it; Vo=sqrt((D^2*g)/(2(costheta)^2*deltaY-2(costheta)^2*tantheta*D)). Could you please find a problem in that equation? (Notice that g is not squared, it was my mistake in the upper post.)

6. Dec 30, 2015

### CWatters

That's where the other inequality come in. You need both... y(6D)>=2D and y(7D)<2D. The first stops undershoot and the second stops overshoot. If both are met the water will go in. I don't think both can be met and the water not go in.

7. Dec 30, 2015

### Fightfish

$y(6D)$ and $y(7D)$ cannot both be the same! The projectile follows a parabolic trajectory. That is why there are two constraints, firstly $y(6D)\geq 2D$ to let the projectile go over the first boundary and the second, $y(7D) < 2D$ to ensure that it does not fly over the second boundary.
Any errors from your trajectory equation carry over directly to your expression for $v_{0}$...
At first glance, you have a sign problem, and some numerical factors are missing. What I got was (using your notation) $$v_{0,min}^{2} = \frac{36D^{2}g}{2\cos^{2}\theta \left(6D \tan \theta - \Delta y\right)},$$
which you can check evaluates to $v_{0,min} = 3\sqrt{Dg}$.

8. Dec 30, 2015

### CherryWine

Yes, you're right for the inequalities. I didn't think about them simultaneously. Sorry. Also Fightfish you are correct with your answer, just checked it both for Vmin and Vmax. My mistake was numerical. Thank you very much.