Projectile Question (1 Viewer)

[Lugia]

A body is projected from a point on a horizontal plane reaches a greatest height h before hitting the plane R metres away from its point of projection. Show that the equation of the body's path may be expressed by y = [4hx(R-x)] / R^2

Somebody PLEASE show me how to do this question!!!! I've been stuck on this since forever

 

[-=«MÄLÅÇhïtÊ»=-]

Have u tried finding equation of trajectory, and equations of range and max height. Should be able to obtain the answer that way. Refer to the answer for a guide towards which direction to go.
The answer isn't in terms of time, so use t as a parameter, to eliminate it.
Play around with the trig equations

 

[Dumbarse]

malachite what u doin next year?
what UAI u aiming for??

 

[SkAnDi]

Ahhh, i tried to solve it... i got pissed off because i couldn't do it at first so perseverance saved the day.

I'm sure there's a more elegant and efficient way to do it though.

But doing it this way does provide.. A solution!

Sorry about the poor quality, i don't have a scanner so took it quickly with my digicam.

EDIT : Just got pointed out to me that... i used Rsin0t instead of Vsin0t... i don't know why it still works though

Well, if you've got a solution i'd like to hear it.. beacuse it's still bugging me how to do it hehe.

 

[Lugia]

hi Skandi, thanks for trying to do this question for me. But I have a question, when u say x = 1/2 R, y = 1/2 [V^2sin^2]/g, ( I changed ur R's into V's... ) how did u get the next line after it?

 

[SkAnDi]

I subbed those values into the y = kx(R-x) to find the value of k.

 

[-=«MÄLÅÇhïtÊ»=-]

dumbarse

not definite
actuary?
or bc of commerce

~97.5 uai

 

[Lugia]

how did u get that equation, y = kx(R-x)? and what's k?

 

[SkAnDi]

And i STILL! don't know why the answer looks right, but i'm meant to use the Vsin0 bit.

All projectile motion in 3u are in the form of an upside down parabola.

The object will be at y=0 (at the ground)
- at the start of the motion
- when x = R (at the range).

Since it's a parabola you know the turning point will be in the middle of the range (1/2R).

So the form of the equation of motion will be something like

y= x(a-x)

When you solve that... the solutiosn will be x=o (yes), and x=a (yes.. in this case it will be R).

So it'll be y= x(a-x).

BUT! It's possible that the equation of motion can contain other constants that are elimitated when you sub those values in (since you'll make it zero). So that's where the k comes in.

You sub the x=1/2r and y = whatever, into the equation to solve for k, so that you can find what the equation of the motion actually is.

 

[Lugia]

hey we never learn that at school!