Calculating Boat Speed in 2-D Relative Motion

[6Stang7]

well, in this problem the smallest max range needed is 12m, but that is for later.

what i mean is when the object has reached its max range it has travled the max distance in the x direction. if you threw a ball, the max range would be the distance from you to where the ball landed (in the x direction).

so when the object has reached its max range, what else do we know? what is the object height when it has reached the max range?

 

[BunDa4Th]

okay, now I am at a lost. I understand what you mean but I don't understand how I would set the equation to find the max range. and to find the max height i believe I would use V_y^2 = V_0y^2 - 2G(deltay)

 

[6Stang7]

ok, let's see if we can make this a little easier.
we have y=C1(R)-C2(R^2).
when the object has reached its max range, it has finished and is at the end of its trajectory. this means that the height is 0, or that y=0.

so if y=0 for max range, then y=0=C1(R)-C2(R^2).
so we now know that the max range of a projectial is C1(R)-C2(R^2)=0.
now solve this equation for R and what do you get?

 

[BunDa4Th]

C1r = C2r^2

C1r/r = C2r^2/r = C1 = C2r

R = C1/c2

 

[6Stang7]

BINGO!
now, re-write that equation by replacing C1 and C2 with what they are.

 

[BunDa4Th]

R = (V_0tan14)/(1.2(9.8))/(V_0cos14)^2)

 

[6Stang7]

ok, now simplfy that equation and what do you get?

 

[BunDa4Th]

R = V_0tan14/1 x 1/2(9.8)/(V_0cos14)^2

1.22V_0/(V_0cos14)^2

R = 1.30V_0

 

[6Stang7]

check you math, you did something wrong.
you should re-work it and get this
R=(((V_0)^2)/9.8)(sin14cos14)

now, do you see how this equation is important?

 

[BunDa4Th]

oh i see what I did wrong i forgot to flip the second equation when multiplying division.

actually wait how did you get that.

R = V_0tan14 x (V_0cos14)^2/1/2(9.8)...

where did 1/2 go

 

[6Stang7]

i wrote it out in microsoft equation to make it easier to understand

the sine14cos14 becomes sin2(14) by the double angle forumla.

i got to go jogging for the night, so i'll be back in about a hour. but do you what this eqaution allows you to solve for?

 

[BunDa4Th]

the link said its invalid.

 

[BunDa4Th]

I would say it will solve for the initial velocity.

 

[BunDa4Th]

okay if i solve it correct V_0 = 22.61

 

[6Stang7]

yes it will solve for initial velocity.
when i did the calculation for it and got 15.83 m/s

let me see if i can get the picture to work.

 

[6Stang7]

the link said its invalid.

hmmm, the link works for me. if it still doesn't work let me know and i can e-mail it to you if you want.

 

[BunDa4Th]

The link still don't work for me. my email address is bunda4th@gmail.com

 

[6Stang7]

i sent it to you. hopefull it will work and help.

 

[BunDa4Th]

it works and i understand it a bit.

Im still a bit confuse how you get sin2(theta) and what does R equal to? or is it just there?

 

[6Stang7]

well, the sin2(theta) came from the sin(theta)cos(theta); it's a trig idenity.
sin(x)cos(x)=sin2(x)

in the final equation we get, R is the range, or the distance travled in the x direction. let's go back to the example i used last time. if i threw a ball and it landed 10 meters away from me, then the range of the ball is 10 meters. range is just another way to say the distance that was travled in the x direction. in the motorcycles case, it needs to just a distance of 12 meters, the smallest range needed is 12 meters.
using algebra you can solve the equation for V_0. in other words, given an initial angle and distance travled (or required to travle) you can calculate the velocity needed.

 

[BunDa4Th]

Oh man you are such a big help in solving this. I now understand it much better. Thanks very much for the time on helping me solve this equation step by step. Again I Thank you

 

[6Stang7]

You're very welcome. :)