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Height and Range of a projectile 
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#1
Jul409, 01:41 PM

P: 14

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] 


#2
Jul409, 03:33 PM

P: 14

1. The problem statement, all variables and given/known data
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. 2. Relevant equations Hmax= [tex]\frac{\left(v0sin\theta\right)^{2}}{2g}[/tex] R = [tex]\frac{v0^{2}sin2\theta}{g}[/tex] 3. 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
Jul409, 04:10 PM

P: 31

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
Jul509, 02:56 PM

Sci Advisor
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Thanks
P: 26,157

Height and Range of a projectile
Hi clarineterr! Welcome to PF!
(have a theta: θ ) 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 sec^{2} = tan^{2} + 1 


#5
Jul509, 07:56 PM

P: 14

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 


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