Projectile motion maximum range problem

In Summary, the projectile will remain airborne for a maximum distance of and the angle at which it will remain airborne is 45 degrees.
  • #1
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2

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 [itex]t = \frac{2v}{g}sin\theta[/itex] .

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

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

Homework Equations


[tex]v=u+at \\
s=ut+\frac{1}{2}gt^2 \\[/tex]

The Attempt at a Solution



Part A:
Horizontal - [itex] vcos\theta[/itex]
Vertical - [itex]vsin\theta[/itex]

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

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 [itex]t=\frac{vsin\theta}{g}[/itex] but as that only gives it until the max height (halfway through) it needs to be times by two and hence [itex] t=\frac{2v}{g}sin\theta[/itex]

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.
 
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  • #2
On C:
At the time of solution, T, y=0, therefore, T satisfies which equation?
 
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  • #3
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)
 
  • #4
If you divided your vertical position by your horizontal position what would you have? (Hint: its a trig function)
 
  • #5
arildno said:
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?

[tex]
0=vsin(\theta)t-\frac{1}{2}gt^2 \\
\frac{1}{2}gt^2=vsin(\theta)t \\
\frac{\frac{1}{2}gt^2}{t}=vsin(\theta) \\
\frac{1}{2}gt=vsin(\theta) \\
gt=2vsin(\theta) \\
t=\frac{2vsin(\theta)}{g} \\
t=\frac{2v}{g}sin(\theta)
[/tex]
 
  • #6
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)
 
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  • #7
Maiq said:
If you divided your vertical position by your horizontal position what would you have? (Hint: its a trig function)

[itex]vtan(\theta)t-\frac{1}{2}gt^2[/itex] ?
 
  • #8
arildno said:
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.

[tex]
\frac{2v^2}{g}sin\theta cos\theta \\ for 0<= \theta <=90
[/tex]
then using the product rule
[tex]
\frac{2v^2}{g}(-sin\theta sin\theta+cos\theta cos\theta)=0 \\
\frac{2v^2}{g}(-sin^2\theta+cos^2\theta)=0 \\
[/tex]
dividing by the fraction on the left then gets
[tex]
-sin^2\theta+cos^2\theta=0 \\
cos^2\theta=sin^2\theta [/tex]
then divide both sides by the cos^2
[tex]
\frac{cos^2\theta}{cos^2\theta}=\frac{sin^2\theta}{cos^2\theta} \\
1=(\frac{sin\theta}{cos\theta})^2 \\
1=tan^2\theta \\
1=tan\theta \\
tan^{-1}(1)=45
[/tex]
 
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  • #9
FaraDazed said:
[itex]vtan(\theta)t-\frac{1}{2}gt^2[/itex] ?

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.
 
  • #10
Maiq said:
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([itex]\frac{dθ}{dt}[/itex])=[itex]\frac{y}{x}[/itex]. Since the angle you start with will be the same as the angle at the end, [itex]\frac{dθ}{dt}[/itex]=0 therefore sin([itex]\frac{dθ}{dt}[/itex])=0. So you end up with

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

then

[itex]\frac{gt}{2vcos(θ)}[/itex]=tan(θ)

t=[itex]\frac{2v}{g}[/itex]sin(θ)

Its a little more complicated but it still works.
 
  • #11
FaraDazed said:
I think I may have done it. Withoutusing that identity though so not sure how succint it is.

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

That's perfectly okay!
But, rewriting:
[tex]Range=\frac{v^{2}}{g}\sin(2\theta)[/tex]
Differentiate, and set equal to zero:
[tex]0=\frac{2v^{2}}{g}\cos(2\theta)[/tex]
That is,
[tex]\cos(2\theta)=0[/tex]
from which the same result follows! :smile:
 
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What is projectile motion maximum range problem?

The projectile motion maximum range problem is a physics problem that involves calculating the maximum horizontal distance traveled by a projectile launched at an angle from a given initial velocity and height, assuming no air resistance.

What are the key factors that affect the maximum range of a projectile?

The key factors that affect the maximum range of a projectile are the initial velocity, the launch angle, and the height from which the projectile is launched. Other factors such as air resistance and wind can also have an impact on the maximum range.

How do you solve a projectile motion maximum range problem?

To solve a projectile motion maximum range problem, you can use the equation R = (v0^2 * sin(2θ)) / g, where R is the maximum range, v0 is the initial velocity, θ is the launch angle, and g is the acceleration due to gravity. You can also use the kinematic equations of motion to solve for the maximum range.

What is the relationship between launch angle and maximum range in projectile motion?

The relationship between launch angle and maximum range in projectile motion is that as the launch angle increases, the maximum range also increases, reaching a maximum at a launch angle of 45 degrees. Beyond 45 degrees, the maximum range decreases as the launch angle increases.

How does air resistance affect the maximum range in projectile motion?

Air resistance can have a significant impact on the maximum range in projectile motion. As air resistance increases, the maximum range decreases due to the opposing force it exerts on the projectile. This can be taken into account by adjusting the initial velocity in the equation for maximum range.

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