Derivitives Help!

  • #1

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.
 

Answers and Replies

  • #2
993
13
X = VT

Vx = V(cosΘ)
Vy = V(sinΘ)

T = d / Vo(sinΘ)
The last equation is not correct.
 
  • #3
HallsofIvy
Science Advisor
Homework Helper
41,833
956
The only acceleration is -g in the y direction, the initial position is (0,0), and the initial velocity is [itex]v_x= V_0 cos(\theta)[/itex], [itex]v_y= V_0 sin(\theta)[/itex]:

[itex]a_x= 0[/itex], [itex]a_y= -g[/itex]

[itex]v_x= V_0 cos(\theta)[/itex], [itex]v_y= -gt+ V_0 sin(\theta)[/itex]

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

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 [itex]tan(\theta)[/itex] or [itex]cot(\theta)[/itex]. The projectile is NOT moving in a straight line so there is no need to look for a "slope".

(y will be largest when [itex]v_y= 0[/itex]. x will be largest when y= 0 again.)
 
  • #4
22
0
(y will be largest when [itex]v_y= 0[/itex]. 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.
 

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