Question

The range (R) of a projectile fired with an initial velocity (Vv0) at an angle theta with the horixontal is: see additional equation
where g is the acceleration due to gravity. Find the angle theta such that the range is a maximum (using calculus)

Answer 1

\[R=(V _{0}\sin2\theta)/g\]

Answer 2

if it helps i know the answer is \[\pi/4\] but i dont know how to get there

Answer 3

@akaisha Do you know what is the angle when the range is maximum ?

Answer 4

no, that is what your supposed to find \[\theta \] somehow using equation they give.

Answer 5

Well we know that the angle when the range is maximum is \[45^{0}\]

Answer 6

I am attaching a graph of R vs sin(2*theta) that I did in Python
>>> x=linspace(0,pi/2)
>>> t=x*180/pi
>>> plot(x,sin(2*x))
the maximum is in the middle between 0,pi/2
That is pi/4

Answer 7

\[ \frac{dR }{ d \theta }=2\frac{ V _{0} }{ g }\cos2\theta\]

Answer 8

\[0=2\frac{ v _{0} }{ g}\cos2\theta\]

Answer 9

so @irene22988 you just take the derivative of the equation given and solve for x and then plug back in for y?

Answer 10

find the derivative then equate it to zero

Answer 11

Ok that seems to make sense. Thanks! :)

Answer 12

no probs

Answer 13

its really easy to prove....

\[\sin2\alpha\] has a maximum value of 1... which is needed to find the maximum range...


so

\[\sin(90) = 1\]

therefore

\[2 \alpha = 90^{o}\]

then \[\alpha = 45^{o}\]

or the launch angle for maximum range is 45 degrees...

Answer 14

i could do it that way.. but we had to use derivatives to find in this section

Answer 15

lol... well then you need to also find the 2nd derivative... as the 1st derivative gives stationary points... and the 2nd derivative is used to test the nature of the stationary points...

max, min or pt of inflexion..

and I think the velocity in your range formula should be squared...just a thought