- #1

- 14

- 0

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]