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Height and Range of a projectile

  • #1
A projectile is fired at a speed v0 from and angle [tex]\theta[/tex] above the horizontal. It has a maximum height H and a range R (on level ground)
Find:
The angle [tex]\theta [/tex] above the horizontal in terms of H and R

The initial speed in terms of H, R and g

and the time of the projectile in terms of H and g.

Relevant Equations:
Hmax= [tex]\frac{\left(v0sin\theta\right)^{2}}{2g}[/tex]
R = [tex]\frac{v0^{2}sin2\theta}{g}[/tex]

Attempt at a solution:

From the maximum height equation: v0sin[tex]\theta[/tex]=[tex]\sqrt{2gh}[/tex]
and from the Range equation: v0cos[tex]\theta[/tex]= [tex]\frac{gR}{2v0sin\theta}[/tex]

then we have v0cos[tex]\theta[/tex]= [tex]\frac{gR}{\sqrt{2gH}}[/tex]

Then tan[tex]\theta[/tex]= [tex]\frac{v0sin\theta}{v0cos\theta}[/tex] = [tex]\frac{2H}{R}[/tex]

so then [tex]\theta[/tex] = tan[tex]^{-1}[/tex][tex]\frac{2H}{R}[/tex]

Then for the second question, I have v0 = [tex]\sqrt{\frac{gR}{sin2\theta}}[/tex]
Then I dont know how to convert it to just be in terms of g, H and R

For the third question I am getting: t = [tex]\frac{2vosin\theta}{g}[/tex]
 

Answers and Replies

  • #2

Homework Statement



A projectile is fired at a speed v0 from and angle [tex]\theta[/tex] above the horizontal. It has a maximum height H and a range R (on level ground)
Find:
The angle [tex]\theta [/tex] above the horizontal in terms of H and R

The initial speed in terms of H, R and g

and the time of the projectile in terms of H and g.

Homework Equations



Hmax= [tex]\frac{\left(v0sin\theta\right)^{2}}{2g}[/tex]
R = [tex]\frac{v0^{2}sin2\theta}{g}[/tex]

The Attempt at a Solution



From the maximum height equation: v0sin[tex]\theta[/tex]=[tex]\sqrt{2gh}[/tex]
and from the Range equation: v0cos[tex]\theta[/tex]= [tex]\frac{gR}{2v0sin\theta}[/tex]

then we have v0cos[tex]\theta[/tex]= [tex]\frac{gR}{\sqrt{2gH}}[/tex]

Then tan[tex]\theta[/tex]= [tex]\frac{v0sin\theta}{v0cos\theta}[/tex] = [tex]\frac{2H}{R}[/tex]

so then [tex]\theta[/tex] = tan[tex]^{-1}[/tex][tex]\frac{2H}{R}[/tex]

Then for the second question, I have v0 = [tex]\sqrt{\frac{gR}{sin2\theta}}[/tex]
Then I dont know how to convert it to just be in terms of g, H and R

For the third question I am getting: t = [tex]\frac{2vosin\theta}{g}[/tex]
 
  • #3
31
0
You know from this equation, Hmax= [tex]\frac{\left(v0sin\theta\right)^{2}}{2g}[/tex], that

v0sin(theta) = sqrt(2Hg)

So plug sqrt(2Hg) into:

t = [tex]\frac{2vosin\theta}{g}[/tex]

to get:

t = 2sqrt(2Hg)/g = 2sqrt(2H/g)
 
  • #4
tiny-tim
Science Advisor
Homework Helper
25,832
251
Welcome to PF!

Hi clarineterr! Welcome to PF! :smile:

(have a theta: θ :wink:)
Find:

The initial speed in terms of H, R and g

and the time of the projectile in terms of H and g.

Then tan[tex]\theta[/tex]= [tex]\frac{v0sin\theta}{v0cos\theta}[/tex] = [tex]\frac{2H}{R}[/tex]

so then [tex]\theta[/tex] = tan[tex]^{-1}[/tex][tex]\frac{2H}{R}[/tex]

For the third question I am getting: t = [tex]\frac{2vosin\theta}{g}[/tex]
Learn your trigonometric identities …

you know tanθ in terms of H R and g, so you simply need to express sin2θ and sinθ in terms of tanθ.

Hint: use sin = cos tan, and cos = 1/sec, and sec2 = tan2 + 1 :wink:
 
  • #5
I got

[tex]\sqrt{\frac{gR^{2}}{4H}\left(\frac{4H^{2}}{R^{2}}+1\right)}[/tex]

??? I dont know if I simplified this right
 
  • #6
tiny-tim
Science Advisor
Homework Helper
25,832
251
For √(gR/sin2θ) ?

Yup, that looks good! :biggrin:

(and now how about your t = 2v0sinθ/g ? :smile:)
 

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