# Height and Range of a projectile

1. ### clarineterr

14
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 = $$\frac{2vosin\theta}{g}$$

2. ### clarineterr

14
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 = $$\frac{2vosin\theta}{g}$$

3. ### MyNewPony

31
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)

4. ### tiny-tim

26,054
Welcome to PF!

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 sec2 = tan2 + 1

5. ### clarineterr

14
I got

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

??? I dont know if I simplified this right

6. ### tiny-tim

26,054
For √(gR/sin2θ) ?

Yup, that looks good!