Height and Range of a projectile

[clarineterr]

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}

[clarineterr]

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}

[MyNewPony]

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)

[tiny-tim]

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\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}

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:

[clarineterr]

I got

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

??? I dont know if I simplified this right

[tiny-tim]

For √(gR/sin2θ) ?

Yup, that looks good! :biggrin:

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