Answer 1

find r'(t) and |v(t)| "This is function of angle theta and t"

Answer 2

@ikram002p @phi

Answer 3

r(t)=<v cos (theta)t+V sin(theta)t-0.5gt^2>
v is diffrent in x and y ?

Answer 4

g is constant ?

Answer 5

my bad no it is \[V0\]

Answer 6

yes

Answer 7

type it again :3

Answer 8

\[<v \cos \theta t,v \sin \theta t-\frac{ 1 }{ 2 } g t^2>\]

Answer 9

The"total" question is at what angle should the projectile be fired to achieve max total distance

Answer 10

this is a position vector with respect to t and theta?

Answer 11

my first thought is maximize
r(t) dot r(t) (square of the distance)

Answer 12

so is it r(t) ?

Answer 13

you can try using partial derivatives with respect to maximize distance as a function of t.

Answer 14

with respect to theta*

Answer 15

my professor gave the answer on how to solve. She said "v(t) = r ' (t) = .....

Find | v(t) |. This is function of angle @ and t.


Now it hits the ground gives t = (2 v/ g) sin @. Actually this gives limits of t for the entire distance travelled from t = 0 to (2v/g) sin @, when it completes its path.


So the distance travelled by the projectile say L(@) , is given by

L(@) = Integral of | v(t) | dt within limits 0 to (2v/g) sin @.
You may use the tables at the back of the book for integration purpose.

IT will come out to bev^2/ g sin @ + v^2/ 2g cos ^2 @ ln [ (1+ sin @)/ (1 - sin @) ]

to maximize L9@) for 2 between 0 and 90 degrees,

L '(@) = 0, gives @ nearly 0.9855.
Compare the L value at this Angle and end points 0 and 90 degress.
Max value of L (1.20 v^2/g) is at this angle ).9855 which is nearly 56 degrees"

Answer 16

I don't understand why we r taking derivative and then taking integral also is this partial derivative?

Answer 17

because you want to maximize the function for t while keeping theta variable...

Answer 18

but this might be a little rough...actually I think I was confused on what your vector represented. Is it the position vector?

Answer 19

yes

Answer 20

do you want max arclength or distance in x direction.

Answer 21

I assume only theta is an argument to cos or sin? not theta*t
in other words, t is multiply cos(theta) ?

Answer 22

yes

Answer 23

\[r(t)=(v \cos \theta t)i+[v \sin \theta t-\frac{ 1 }{ 2 }g t^2]j\]

Answer 24

from the knowledge of physics g= gravity and t = time

Answer 25

it's less confusing if you write it as
\[ r(t)=(v \ t \cos \theta )i+[v \ t \sin \theta -\frac{ 1 }{ 2 }g t^2]j \]

Answer 26

well since its r(t) then it wud be one variable fun :-\
idk theta situation sounds weird xD

Answer 27

what do you mean by maximum total distance ?

is this arc length or horizontal distance ?

Answer 28

the reason you need to take the derivative of the integral is the total distance is modeled by the the integral of the speed. Speed is derivative of the norm. so you would use the arc length formula. (also I am confused about the variables as well)

Answer 29

so \[v(t)= <v \cos \theta, v \sin \theta - g t>\]

Answer 30

this is the velocity vector.

Answer 31

ya

Answer 32

first clarify if you mean arc length vs horizontal distance please

Answer 33

I dont know what professor meant by total distance

Answer 34

Integral of | v(t) | dt within limits 0 to (2v/g) sin @.

Answer 35

thats defintely arclength formula....so you need to first find the derivative of the position vector and get the norm of that. This is the speed. next you integrate that with respect to t to get arclength formula.

Answer 36

I think from there you can maximize the function with respect to theta...but I am not sure.

Answer 37

\[\int\limits_{0}^{\frac{ 2v \sin \theta }{ g }}\sqrt{(vsin \theta)^2+(vsin \theta-g t)^2}\]

Answer 38

right?

Answer 39

instead of sin, the first one is cos

Answer 40

lol yes thx, \[v^2/ g \sin \theta + v^2/ 2g \cos ^2 \theta \ln [ (1+ \sin \theta)/ (1 - \sin \theta) ]\]

Answer 41

i need to take derivative of this and find critical points?

Answer 42

What is whole point of this I mean why are we useing arc length

Answer 43

because you prof told you to take integral of norm of derivative, which is the formula for arc length.

Answer 44

so i got 0.98551474 how to get the angle?

Answer 45

you found theta in radians. Convert to degrees (multiply by 180/pi)

sylbot outlined what you are doing:

your position vector
\[ r(t)=(v\ t\cos \theta )i+[v \ t\sin \theta-\frac{ 1 }{ 2 }g t^2]j \]
traces out an arc.
the velocity at any time t is r'(t)
\[ r'(t)= <- v \sin \theta , \ v \cos \theta - g t > \]
which is a vector with magnitude and direction
The speed is the magnitude of the velocity vector

If the speed were constant, the distance traveled would be speed * time
but when the speed s(t) varies we integrate
\[ \int s(t) dt \]

Answer 46

to find the bounds for the integral, we assume the arc begins and ends at height zero, or in other words, with the y-component of the position vector= 0
solving
\[ v \ t\sin \theta-\frac{ 1 }{ 2 }g t^2 = 0 \]
for t we get t=0 and t= (2v/g) sin theta

Answer 47

after integrating you have found the length of the arc as a function of the launch angle theta. Take the derivative with respect to theta to find the critical points (max in this case) of the length of the arc.
you find theta= 0.98551474 radians or 56.47 degrees

Answer 48

if you use theta= 0.98551474 in the formula for the length of the arc
\[ \frac{v^2}{ g} \sin \theta + \frac{v^2}{2 g} \cos ^2 \theta \ \ln [ (1+ \sin \theta)/ (1 - \sin \theta) ] \]
and simplify , you get
\[ 1.1996 \frac{v^2}{ g} \]

compare that to using theta= 0 or theta=pi/2