Derivatives Help: Projectile Fired w/ Initial Speed Vo

[Gabriel1234]

Homework Statement

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

Homework Equations

X = VT

Vx = V(cosΘ)
Vy = V(sinΘ)

T = d / Vo(sinΘ)

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.

 

[grzz]

X = VT

Vx = V(cosΘ)
Vy = V(sinΘ)

T = d / Vo(sinΘ)

The last equation is not correct.

 

[HallsofIvy]

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.)

 

[GrantB]

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

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.