# Derivitives Help!

1. Sep 28, 2011

### Gabriel1234

1. The problem statement, all variables and given/known data

A projectile is fired with initial speed Vo at an angle Θ above the horizontal over flat ground.

a. show that the distance d that the projectile goes and height h reached by the projectile are given by:

d = (Vo^2 sin2Θ)/g
h = (Vo^2 sin^2Θ)/2g

2. Relevant equations

X = VT

Vx = V(cosΘ)
Vy = V(sinΘ)

T = d / Vo(sinΘ)

3. The attempt at a solution

d = (Vo^2 sin2Θ)/g
Vt = (Vo^2 sin2Θ)/g
(Vocos(Θ))(d / Vosin(Θ)) = (Vo^2 sin2Θ)/g

dcot(Θ) = (Vo^2 sin(2Θ)/g
dcot(Θ) = (Vo^2 * 2sin(Θ)cos(Θ))/g

Then this is where I get lost.

If I do divide cotangent of theta then would distance be proved.
I haven't started on height yet.

2. Sep 29, 2011

### grzz

The last equation is not correct.

3. Sep 29, 2011

### HallsofIvy

Staff Emeritus
The only acceleration is -g in the y direction, the initial position is (0,0), and the initial velocity is $v_x= V_0 cos(\theta)$, $v_y= V_0 sin(\theta)$:

$a_x= 0$, $a_y= -g$

$v_x= V_0 cos(\theta)$, $v_y= -gt+ V_0 sin(\theta)$

$x= V_0 cos(\theta)t+ 0$, $-(g/2)t^2+ V_0 sin(\theta)+ 0$.

Since the problem asks for the x and y distances separately, I see no reason to combine x and y and so no reason to look at $tan(\theta)$ or $cot(\theta)$. The projectile is NOT moving in a straight line so there is no need to look for a "slope".

(y will be largest when $v_y= 0$. x will be largest when y= 0 again.)

4. Sep 29, 2011

### GrantB

Yes. Think about it like this: What is the slope of the parabola at its maximum?

It appears as though you are just trying to throw equations together, and work backwards, to get the answer. Take a minute and think about the problem.