Projectile motion maximum range problem

[FaraDazed]

Homework Statement

A projectile is fired at 30 km/h at the angle θ in x direction of an x-y plane. Ignoring the effects of air resistance, find the maximum
range and the value of θ that gives this range

Part A: write down the initial horizontal and vertical components
of the velocity

Part B: Write two equations to give the horizontal and vertical positions of the
projectile as functions of time in terms of the initial speed, and the initial
angle .

Part C: Use these equations to show that the time the projectile will remain
airborne is given by t = \frac{2v}{g}sin\theta .

Part D: Hence show that the range is given by \frac{2v^2}{g}sin(\theta)cos(\theta)

Part E: Show, by differentiating, that the maximum horizontal distance will occur if fired at 45° and is given by x_{max}=\frac{v^2}{g}

Homework Equations

v=u+at \\<br /> s=ut+\frac{1}{2}gt^2 \\

The Attempt at a Solution

Part A:
Horizontal - vcos\theta
Vertical - vsin\theta

Part B:
Horizontal - vcos(\theta)t
Vertical - vsin(\theta)t+\frac{1}{2}gt^2

Part C is where I am stuck. I understand it but when I done it before I did it a different way (i think).

When I did it before (last year) I used v=at so t=v/a so t=\frac{vsin\theta}{g} but as that only gives it until the max height (halfway through) it needs to be times by two and hence t=\frac{2v}{g}sin\theta

But the question says to use what I did in Part B and I am not sure how that is done.

Part D: that is straightforward, just substituting what is found as t in Part C into the horizontal bit of Part B.

Part E: This part has my head spinning at the moment.

 

[arildno]

On C:
At the time of solution, T, y=0, therefore, T satisfies which equation?

 

[arildno]

E:
You are to find maximal range. The range is given as a function of the angle v.
Hint:
It is easiest to first rewrite the expression for the range using the trig identity sin(v)cos(v)=1/2sin(2v)

 

[Maiq]

If you divided your vertical position by your horizontal position what would you have? (Hint: its a trig function)

 

[FaraDazed]

On C:
At the time of solution, T, y=0, therefore, T satisfies which equation?

OK I think I see what you mean, like this?

<br /> 0=vsin(\theta)t-\frac{1}{2}gt^2 \\<br /> \frac{1}{2}gt^2=vsin(\theta)t \\<br /> \frac{\frac{1}{2}gt^2}{t}=vsin(\theta) \\<br /> \frac{1}{2}gt=vsin(\theta) \\<br /> gt=2vsin(\theta) \\<br /> t=\frac{2vsin(\theta)}{g} \\<br /> t=\frac{2v}{g}sin(\theta)<br />

 

[arildno]

That's absolutely right!

General cautionary note:
Note that when you divided by "t", you lost one solution, namely t=0 (that is of course, the INITIAL y-position, so it isn't really relevant here, but you should not make a habit to divide with a quantity that MIGHT be 0)

 

[FaraDazed]

If you divided your vertical position by your horizontal position what would you have? (Hint: its a trig function)

vtan(\theta)t-\frac{1}{2}gt^2 ?

 

[FaraDazed]

E:
You are to find maximal range. The range is given as a function of the angle v.
Hint:
It is easiest to first rewrite the expression for the range using the trig identity sin(v)cos(v)=1/2sin(2v)

I think I may have done it. Withoutusing that identity though so not sure how succint it is.

<br /> \frac{2v^2}{g}sin\theta cos\theta \\ for 0&lt;= \theta &lt;=90<br />
then using the product rule
<br /> \frac{2v^2}{g}(-sin\theta sin\theta+cos\theta cos\theta)=0 \\<br /> \frac{2v^2}{g}(-sin^2\theta+cos^2\theta)=0 \\<br />
dividing by the fraction on the left then gets
<br /> -sin^2\theta+cos^2\theta=0 \\<br /> cos^2\theta=sin^2\theta
then divide both sides by the cos^2
<br /> \frac{cos^2\theta}{cos^2\theta}=\frac{sin^2\theta}{cos^2\theta} \\<br /> 1=(\frac{sin\theta}{cos\theta})^2 \\<br /> 1=tan^2\theta \\<br /> 1=tan\theta \\<br /> tan^{-1}(1)=45<br />

 

[Maiq]

vtan(\theta)t-\frac{1}{2}gt^2 ?

Never mind, this equation doesn't seem to work. What I was trying to do was use sin(θ)=y/x. I now realize that the θ in this sin(θ) changes with time, while the angle θ is constant in the position equations.

 

[Maiq]

Never mind, this equation doesn't seem to work. What I was trying to do was use sin(θ)=y/x. I now realize that the θ in this sin(θ) changes with time, while the angle θ is constant in the position equations.

Ok so it does work actually but you would have to use sin(\frac{dθ}{dt})=\frac{y}{x}. Since the angle you start with will be the same as the angle at the end, \frac{dθ}{dt}=0 therefore sin(\frac{dθ}{dt})=0. So you end up with

0=\frac{vsin(θ)t-(1/2)gt^{2}}{vcos(θ)t}=tan(θ)-\frac{gt}{2vcos(θ)}

then

\frac{gt}{2vcos(θ)}=tan(θ)

t=\frac{2v}{g}sin(θ)

Its a little more complicated but it still works.

 

[arildno]

I think I may have done it. Withoutusing that identity though so not sure how succint it is.

<br /> \frac{2v^2}{g}sin\theta cos\theta \\ for 0&lt;= \theta &lt;=90<br />
then using the product rule
<br /> \frac{2v^2}{g}(-sin\theta sin\theta+cos\theta cos\theta)=0 \\<br /> \frac{2v^2}{g}(-sin^2\theta+cos^2\theta)=0 \\<br />
dividing by the fraction on the left then gets
<br /> -sin^2\theta+cos^2\theta=0 \\<br /> cos^2\theta=sin^2\theta
then divide both sides by the cos^2
<br /> \frac{cos^2\theta}{cos^2\theta}=\frac{sin^2\theta}{cos^2\theta} \\<br /> 1=(\frac{sin\theta}{cos\theta})^2 \\<br /> 1=tan^2\theta \\<br /> 1=tan\theta \\<br /> tan^{-1}(1)=45<br />

That's perfectly okay!
But, rewriting:
Range=\frac{v^{2}}{g}\sin(2\theta)
Differentiate, and set equal to zero:
0=\frac{2v^{2}}{g}\cos(2\theta)
That is,
\cos(2\theta)=0
from which the same result follows!