Help with identities and projectiles

[OhMrsDarcy]

The problem states:

The range of a projectile is defined as the horizontal distance it travels in flight. If the path a projectile with initial velocity Vo and initial angle of inclination thetao is given by:

. . .y = (g/2Vo) sec^2 (Thetao) x^2+(tan(Thetao))x

Find the angle of inclination that gives the maximum range. Justify your answer.

Sorry: I don't know how to make the actual Greek letter come up for theta, or how to make the squares look like they're being squared. But i could really use some help with this one

 

[stapel]

Just to clarify, is the following your function?

. . . . .\(\displaystyle \L y\, =\, \left(\frac{g}{2V_0}\right)\, \left(\sec^2{(\theta_0)}\right)\,x^2\,+\,\left(\tan{(\theta_0)}\right)\,x\)

Thank you.

Eliz.

 

[soroban]

Hello, OhMrsDarcy!

The range of a projectile is defined as the horizontal distance it travels in flight.
If the path a projectile with initial velocity \(\displaystyle v\) and initial angle of inclination \(\displaystyle \theta\) is given by:

. . .\(\displaystyle \L y \:= \:\frac{g}{2v}(\sec^2\theta)x^2\,+\,(\tan\theta)x\)

Find the angle of inclination that gives the maximum range.

The projectile is on the ground when \(\displaystyle y \,= \,0\)

We have: \(\displaystyle \L\:\frac{g}{2v}(\sec^2\theta)x^2 \,+\,(\tan\theta)x \;= \;0\)

Factor: \(\displaystyle \L\:x\left[\frac{g}{2v}(\sec^2\theta)x \,+\,\tan\theta\right] \;= \;0\)


Set each factor equal to 0 and solve.

\(\displaystyle \L\;\;x\,=\,0\) . . . Of course it's on the ground at the very beginning.

\(\displaystyle \L\;\;\frac{g}{2v}(\sec^2\theta)x\,+\,\tan\theta \:=\:0\;\;\Rightarrow\;\;x \:=\:-\frac{\tan\theta}{\frac{g}{2v}\sec^2\theta} \:=\:-\frac{2v}{g}\tan\theta\cos^2\theta\)

and we have: \(\displaystyle \L\:x\:=\:-\frac{2v}{g}\sin\theta\cos\theta \:=\:-\frac{v}{g}\sin(2\theta)\)


The range \(\displaystyle (x)\) is a maximum when \(\displaystyle \,\sin(2\theta) \,= \,1\)

Therefore, we have: \(\displaystyle \L\,2\theta \,=\,90^o\;\;\Rightarrow\;\;\fbox{\theta\,=\,45^o}\)