Projectile maths

[jongeeone]

Hi, just looking for some quick help!

Been trying to do this for ages and can't seem to work it out. It's a projectile problem that ignores all resistance etc and models the projectile as a particle so it has no mass.

The projectile leaves the ground at 45 degrees at a speed of 4.47. Need to work out the time before it reaches maximum height, and what height this is so i can determine if the particle will hit the target (a windscreen) at 1 meter high.

Any help/input appreciated, thanks.

 

[FoxEye]

Mechanics is like some kind of voodoo magic, practiced by alien wizards.

Statistics and pure maths I could handle; mechanics I flunked :/

 

[Nate--IRL--]

Split the problem into X and Y vectors?

Nate

 

[Alex74]

let y be vertical displacement, x be horizontal displacement.
(d^2)y/d(t^2) = -9.1, (d^2)x/d(t^2) = 0

=> dy/dt = -9.1t + c, dx/dt = h

plug initial values, c = 4.47*sin(45), h = 4.47*cos(45)

Then solve dy/dt = 0. That'll be at some time, call it t0.

integrate again,

y = -4.55*t^2 + c*t + c1 , c1 a new constant,

x = h*t + h1, h1 a new constant


solve for y = 0 to get a time t1, plug t1 into x, t0 into y and you're done

EDIT: 9.1 should be -9.1...my bad *changed*

 

[Dunks]

let y be vertical displacement, x be horizontal displacement.
(d^2)y/d(t^2) = 9.1, (d^2)x/d(t^2) = 0
=> dy/dt = 9.1t + c, dx/dt = h

plug initial values, c = 4.47*sin(45), h = 4.47*cos(45)

Then solve dy/dt = 0. That'll be at some time, call it t0.

integrate again,

y = 4.55*t^2 + c*t + c1 , c1 a new constant,

x = h*t + h1, h1 a new constant


solve for y = 0 to get a time t1, plug t1 into x, t0 into y and you're done

Talk about over complicating things .

Max height = 50.94cm and time taken is 0.64 seconds.

 

[Hxc]

SUVAT with the vertical speed. Max height is when V=0
S=?
U=4.47Sin45
V=0
A=-9.98
T= unimportant

use v² = u² + 2as

and you'll work out the max height, from which you can get the time

Mech1 was a long time ago and I'm tired. there is a simpler way of doing this by calculating time first. i am dumb.

 

[digipeep]

Get a iPhone and play Angry Birds, you'll sail your CSE Maths no problem

 

[Alex74]

Talk about over complicating things .

Max height = 50.94cm and time taken is 0.64 seconds.

maybe, but it's the most intuitive method imo, plus it's like..5 lines

 

[Amp34]

let y be vertical displacement, x be horizontal displacement.
(d^2)y/d(t^2) = -9.1, (d^2)x/d(t^2) = 0

=> dy/dt = -9.1t + c, dx/dt = h

plug initial values, c = 4.47*sin(45), h = 4.47*cos(45)

Then solve dy/dt = 0. That'll be at some time, call it t0.

integrate again,

y = -4.55*t^2 + c*t + c1 , c1 a new constant,

x = h*t + h1, h1 a new constant


solve for y = 0 to get a time t1, plug t1 into x, t0 into y and you're done

EDIT: 9.1 should be -9.1...my bad *changed*

And this post shows that knowing the principles behind the maths is much better than just knowing the maths!

 

[Dunks]

maybe, but it's the most intuitive method imo, plus it's like..5 lines

I think asking someone who can't figure this out to solve a second order diff. eqn. might be asking a bit much? . Or not going by FoxEyes post .

 

[Amp34]

Integration, hated it with a passion...

e^x isn't too bad though...

 

[Alex74]

Well, hopefully at least one method will prove useful

Can't say i disagree with FoxEye, I'm so glad i'm not forced to do classical mechanics any more . One course in first year was more than enough for me..

 

[Dunks]

Well, hopefully at least one method will prove useful

Can't say i disagree with FoxEye, I'm so glad i'm not forced to do classical mechanics any more . One course in first year was more than enough for me..

Second year Mechanical Engineering here. I'm starting to get pretty resentful towards it now .

 

[Alex74]

Second year Mechanical Engineering here. I'm starting to get pretty resentful towards it now .

2nd year maths . Only real applied work this year was stats (exam for which is today *cries a little*)..

 

[Nimble]

let y be vertical displacement, x be horizontal displacement.
(d^2)y/d(t^2) = -9.1, (d^2)x/d(t^2) = 0

=> dy/dt = -9.1t + c, dx/dt = h

plug initial values, c = 4.47*sin(45), h = 4.47*cos(45)

Then solve dy/dt = 0. That'll be at some time, call it t0.

integrate again,

y = -4.55*t^2 + c*t + c1 , c1 a new constant,

x = h*t + h1, h1 a new constant


solve for y = 0 to get a time t1, plug t1 into x, t0 into y and you're done

EDIT: 9.1 should be -9.1...my bad *changed*

Should be -9.8 actually . Should be solving with y=0, t=0 as well to get c1 and h1.

Easiest way to solve it is with (same as how Alex74 did it really):

v = v0 + at (solve for time)
s = vt + 1/2at^2 (solve for height)

Comes out at t = 0.323s, s = 0.510m

 

[Inquisitor]

let y be vertical displacement, x be horizontal displacement.
(d^2)y/d(t^2) = -9.1, (d^2)x/d(t^2) = 0

=> dy/dt = -9.1t + c, dx/dt = h

plug initial values, c = 4.47*sin(45), h = 4.47*cos(45)

Then solve dy/dt = 0. That'll be at some time, call it t0.

integrate again,

y = -4.55*t^2 + c*t + c1 , c1 a new constant,

x = h*t + h1, h1 a new constant


solve for y = 0 to get a time t1, plug t1 into x, t0 into y and you're done

EDIT: 9.1 should be -9.1...my bad *changed*

All the cool kids use Lagrangian mechanics. Newtonian mechanics is so 17th century.

 

[Alex74]

Should be -9.8 actually ..

oopsie

All the cool kids use Lagrangian mechanics. Newtonian mechanics is so 17th century.

 

[jongeeone]

Thanks for the help, having no luck tho

Tried putting the SUVAT equations into excel and i'm getting no luck.

 

[Strife212]

Mechanics

 

[House]

Back when I did this stuff we were taught 2 diffrent ways to do the same calcs.

A-level physics was all equations of motion (pretty simple stuff)

A-level Maths was all vectors (again simple but not as simple as using the simple equations)

Its all fairly basic stuff, thought it reminds me how little I acutually use any on the maths I know/knew day to day anymore. I guess its good to have an understanding of the calcs that the modeling and simulation packages are running though.