Question

what is the angle that a projectile should be projected at to get the maximum range if you ignore air resistance

Answer 1

should we assume a general velocity; and therefore 1 would be fine?

Answer 2

not 1 degree, 1 m/s :)

Answer 3

This is one of my favorites.

Answer 4

my thought is:
\[\Delta y=-\frac12gt^2+v_osin\theta ~t\]
\[v=-gt+v_osin\theta=0~:~t=\frac{v_osin\theta}{g}\]
\[\Delta x=v_ocos\theta~t\]
\[\Delta x=v_ocos\theta~\frac{v_osin\theta}{g}\]since v_0 is arbitrary; lets use 1
\[\Delta x=\frac{sin\theta ~cos\theta}{g}\]

Answer 5

taking the derivative of delta x and finding its value at 0 is what im thinking next :)

Answer 6

\[\frac1g(cos^2\theta-sin^2\theta)=0~when~cos^2\theta=sin^2\theta\]

Answer 7

what can we gather from this?

Answer 8

..and, did i do it right?

Answer 9

You'll get the right answer.

Might I offer another approach that detours around the Calculus for the case that Rach1125 isn't taking Calculus based Physics.

Let's develop an expression for vertical displacement, \[y = vt \sin(\theta) - {1 \over 2} g t^2\]and for horizontal displacement\[x = vt \cos(\theta)\]

We assume that the object stops travelling when it returns back to earth, y = 0. Let's solve the vertical displacement equation for t\[t = {2 v \sin(\theta) \over g}\]
Let's substitute this back into our horizontal displacement equation: \[x = {2v^2 \cos(\theta) \sin(\theta) \over g}\]This can be simplified with the following trigonometric identity: \[\sin(2\theta) = 2 \sin(\theta) \cos(\theta)\]Applying the identity yields: \[x = {v^2 \sin(2\theta) \over g}\]
We need the maximum of \(\sin(2\theta)\). Since \(\sin(90) = 1\), \(\sin(2*45) = 1\)

Therefore, it can be concluded that the maximum value of x occurs when \(\theta = 45\).

Answer 10

@eashmore: I never really thought of trying to prove it...it's just always made logical sense that it would be 45 degrees. Thanks for the detailed post...good stuff :)

Answer 11

It's always nice to be able to verify our observations/logical inferences with the Physics.

As you might also suspect, if the launching point is ABOVE the landing point, our "maximum range angle" is less than 45; if the launching point if BELOW the landing point, our "maximum range angle" is greater than 45. These proofs require a fair bit of Calculus and Trigonometry.