## Height and Range of a projectile

A projectile is fired at a speed v0 from and angle $$\theta$$ above the horizontal. It has a maximum height H and a range R (on level ground)
Find:
The angle $$\theta$$ 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= $$\frac{\left(v0sin\theta\right)^{2}}{2g}$$
R = $$\frac{v0^{2}sin2\theta}{g}$$

Attempt at a solution:

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

then we have v0cos$$\theta$$= $$\frac{gR}{\sqrt{2gH}}$$

Then tan$$\theta$$= $$\frac{v0sin\theta}{v0cos\theta}$$ = $$\frac{2H}{R}$$

so then $$\theta$$ = tan$$^{-1}$$$$\frac{2H}{R}$$

Then for the second question, I have v0 = $$\sqrt{\frac{gR}{sin2\theta}}$$
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 = $$\frac{2vosin\theta}{g}$$
 1. The problem statement, all variables and given/known data A projectile is fired at a speed v0 from and angle $$\theta$$ above the horizontal. It has a maximum height H and a range R (on level ground) Find: The angle $$\theta$$ 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= $$\frac{\left(v0sin\theta\right)^{2}}{2g}$$ R = $$\frac{v0^{2}sin2\theta}{g}$$ 3. The attempt at a solution From the maximum height equation: v0sin$$\theta$$=$$\sqrt{2gh}$$ and from the Range equation: v0cos$$\theta$$= $$\frac{gR}{2v0sin\theta}$$ then we have v0cos$$\theta$$= $$\frac{gR}{\sqrt{2gH}}$$ Then tan$$\theta$$= $$\frac{v0sin\theta}{v0cos\theta}$$ = $$\frac{2H}{R}$$ so then $$\theta$$ = tan$$^{-1}$$$$\frac{2H}{R}$$ Then for the second question, I have v0 = $$\sqrt{\frac{gR}{sin2\theta}}$$ 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 = $$\frac{2vosin\theta}{g}$$
 You know from this equation, Hmax= $$\frac{\left(v0sin\theta\right)^{2}}{2g}$$, that v0sin(theta) = sqrt(2Hg) So plug sqrt(2Hg) into: t = $$\frac{2vosin\theta}{g}$$ to get: t = 2sqrt(2Hg)/g = 2sqrt(2H/g)

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## Height and Range of a projectile

Hi clarineterr! Welcome to PF!

(have a theta: θ )
 Quote by clarineterr 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$$\theta$$= $$\frac{v0sin\theta}{v0cos\theta}$$ = $$\frac{2H}{R}$$ so then $$\theta$$ = tan$$^{-1}$$$$\frac{2H}{R}$$ … For the third question I am getting: t = $$\frac{2vosin\theta}{g}$$
 I got $$\sqrt{\frac{gR^{2}}{4H}\left(\frac{4H^{2}}{R^{2}}+1\right)}$$ ??? I dont know if I simplified this right