Projectile Problem: Finding Max Range & Height, Time to Reach Ground

[John 123]

Homework Statement

A projectile is fired from a height of
<br /> y_0<br />
ft above a level terrain,with a velocity of
<br /> v_0<br />
and at an angle
<br /> \alpha<br />
with the horizontal.
Find
a. The equation of the particle's path.
I have solved this:
<br /> y=y_0+(\tan\alpha)x-\frac{g(\sec^2\alpha)}{2(v_0)^2}x^2<br />
b.It's horizontal range-take the x-axis on ground level.
I have also solved this as the horizontal range is found from setting y=0
this leads to a quadratic in x and taking the positive root gives the range.
c. It's maximum height.
Again I have solved this by setting dy/dx =0
Finally
<br /> y_{max}=y_0+\frac{v_0^2\sin^2\alpha}{2g}<br />
d. When will it reach the ground?
Again I have solved this.
<br /> t=\frac{range}{v_0\cos\alpha}<br />
The denominator is the horizontal component of vo.
IT IS THE NEXT PART THAT I CANNOT SOLVE.
e. What is the value of alpha that will make the range a maximum?
John

Homework Equations

The Attempt at a Solution

 

[tiny-tim]

Hi John!

Find
a. The equation of the particle's path.
I have solved this:
<br /> y=y_0+(\tan\alpha)x-\frac{g(\sec^2\alpha)}{2(v_0)^2}x^2<br />
b.It's horizontal range-take the x-axis on ground level.
I have also solved this as the horizontal range is found from setting y=0
this leads to a quadratic in x and taking the positive root gives the range.
…
IT IS THE NEXT PART THAT I CANNOT SOLVE.
e. What is the value of alpha that will make the range a maximum?

ok, what is the range that you found, and that you now have to maximise?

 

[kuruman]

In part (d) what is "range"? The given quantities are v₀, α and y₀. Unless you assume that the projectile returns to the same height y₀, you will not be able to proceed. If this assumption is valid, take y₀=0 and proceed as tiny-tim suggests. But you do need to find an expression for the range in terms of the given quantities.

 

[John 123]

OK the range is obtained by setting y=0.
This leads to the following quadratic in x.
<br /> \frac{g\sec^2\alpha}{2v_0^2}x^2-(\tan\alpha)x-y_0=0<br />
This leads to a complicated expression in x the positive root of which is the range.
The problem now is finding
<br /> \frac{dx}{d\alpha}<br />
and setting it to zero to find the value of alpha to make x a maximum.
John

 

[kuruman]

First off set y₀ = 0. Then re-express all trig functions with sines and cosines. Then it should be obvious to you what's going on.

 

[John 123]

With yo=0
<br /> x=\frac{v_0^2}{g}\sin2\alpha<br />
Then with
<br /> \frac{dx}{d\alpha}=0<br />
<br /> \frac{2v_0^2}{g}\cos2\alpha=0<br />
This gives
<br /> \alpha=\pi/4<br />
However how do we solve this problem to include y0.
The book answer to this question is:
<br /> \sin\alpha=\frac{v_0}{\sqrt(2(v_0^2+gy_0))}<br />

 

[kuruman]

First you need to solve the equation you have with y=0
<br /> <br /> 0=y_0+(\tan\alpha)x-\frac{g(\sec^2\alpha)}{2(v_0)^2}x^2<br /> <br />

for x. The larger of the two values is the range (Why?). Then, as tiny-tim said, maximize that function.

 

[tiny-tim]

Hi kuruman!

First off set y₀ = 0. Then re-express all trig functions with sines and cosines. Then it should be obvious to you what's going on.

I think you meant set y = 0.

But using sec² = tan² + 1 is probably easier.

 

[John 123]

I have set y=0
Then the equation is:
<br /> \frac{g\sec^2\alpha}{2v_0^2}x^2-(\tan\alpha)x-y_0=0<br />
Now need to express x explicitly in terms of alpha, then find dx/d(alpha)
then equate
<br /> \frac{dx}{d\alpha}=0<br />
The equation is a quadratic in x and the roots are complicated and finding dx/d(alpha) is the problem?
John

 

[tiny-tim]

The equation is a quadratic in x and the roots are complicated …

… and the day is long , and the wages are small …

get on with it! ​

 

[raknath]

Y₀ is the initial distance? from what?

what is y₀

 

[John 123]

<br /> y_0<br />
is the height above ground level from which the projectile is fired.
John