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VCE Stuff => VCE Science => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Physics => Topic started by: naved_s9994 on April 19, 2009, 01:04:21 pm

Title: Projectile motion formulas
Post by: naved_s9994 on April 19, 2009, 01:04:21 pm: Hey VN Members, there are formulas used for Projectile motion, for landing at same height

Eg. Time to reach max height
Range
Maximum height reached

and many others, if you can please post the formulas than big help, thanks
Title: Re: Projectile motion formulas
Post by: Flaming_Arrow on April 19, 2009, 01:07:07 pm: Range = $\frac{u^2 \sin (2\theta)}{g}$
Title: Re: Projectile motion formulas
Post by: naved_s9994 on April 19, 2009, 01:10:40 pm: ok thanks
Title: Re: Projectile motion formulas
Post by: Damo17 on April 19, 2009, 01:10:45 pm: Quote from: naved_s9994 on April 19, 2009, 01:04:21 pm
Hey VN Members, there are formulas used for Projectile motion, for landing at same height

Eg. Time to reach max height
Range
Maximum height reached

and many others, if you can please post the formulas than big help, thanks

time to reach max height.
$V_v=U_v+at$
solve for $t$

range
$U_h \times t= Range$
input $U_h$ and $t$

max height reached
$(V_v)^2=(U_v)^2+2ax$
solve for $x$
Title: Re: Projectile motion formulas
Post by: naved_s9994 on April 19, 2009, 01:19:14 pm: THANKKSS DAMO 17 HUGE AID !!
AND FLAMING....

AWESOME GUYS !!
Title: Re: Projectile motion formulas
Post by: naved_s9994 on April 19, 2009, 01:23:12 pm: but what components do we use for max height, and time reached then?

thnx
Title: Re: Projectile motion formulas
Post by: Damo17 on April 19, 2009, 01:35:57 pm: Quote from: naved_s9994 on April 19, 2009, 01:23:12 pm
but what components do we use for max height, and time reached then?

thnx

for time to reach max height you would have been given $V$ and $\theta$ or $U_v$ so if given $V$ and $\theta$ work out $U_v= V sin \theta$
$V_v$ at the top is $0$ . $a=-10$ , put these value in and solve for t.

By the way $U_v$ means initial vertical velocity and $U_h$ means initial horizontal velocity. $V_v$ means final vertical velocity.
For max height reached use $V_v=0$ , $U_v=V sin \theta$ and $a=-10$ and solve for $x$ .
Title: Re: Projectile motion formulas
Post by: TrueTears on April 20, 2009, 02:21:36 pm: Ok these are the complete formulas:

For FULL parabolic path:

(' $u$ ' stands for initial speed not the horizontal or vertical.)

$\mbox{Total time of flight}$ $= \frac{2usin\theta}{g}$
$\mbox{Height_{max}}$ $= \frac{u^2sin^2 \theta}{2g}$
$\mbox{Range}$ $= \frac{u^2sin(2 \theta)}{g}$

For HALF parabolic path:

( $'u'$ stands for horizontal speed)

$\mbox{Total time of flight = \sqrt{\frac{2h}{g}}}$
$\mbox{V{vertical speed} = \sqrt{2gh}}$
$\mbox{Range = u\sqrt{\frac{2h}{g}}}$
$v_{\mbox{just before touching ground}} = \sqrt{u^2+2gh}$
Title: Re: Projectile motion formulas
Post by: naved_s9994 on April 20, 2009, 06:42:14 pm: tnx every1 !!
Title: Re: Projectile motion formulas
Post by: dejan91 on April 20, 2009, 10:23:43 pm: Honestly, all you need to remember are the five or so orignial motion equations. Those complicated ones like $\frac{2usin\theta}{g}$ and $\frac{u^2sin^2\theta}{2g}$ can be derived using the original ones.
Title: Re: Projectile motion formulas
Post by: naved_s9994 on April 21, 2009, 04:45:52 pm: Okay, Thanks every, Awesome....Its all clear now....

Once you know the basic formulas, you can answer alll questions much easily :D
Title: Re: Projectile motion formulas
Post by: naved_s9994 on May 12, 2009, 06:38:59 pm: Truetears....For HALF parabloic 'the velocity before touching ground'....umm, ive found the speed....but how would you work out the angle at which it touches the floor?

THANKS

Quote from: TrueTears on April 20, 2009, 02:21:36 pm
Ok these are the complete formulas:

For FULL parabolic path:

(' $u$ ' stands for initial speed not the horizontal or vertical.)

$\mbox{Total time of flight}$ $= \frac{2usin\theta}{g}$
$\mbox{Height_{max}}$ $= \frac{u^2sin^2 \theta}{2g}$
$\mbox{Range}$ $= \frac{u^2sin(2 \theta)}{g}$

For HALF parabolic path:

( $'u'$ stands for horizontal speed)

$\mbox{Total time of flight = \sqrt{\frac{2h}{g}}}$
$\mbox{V{vertical speed} = \sqrt{2gh}}$
$\mbox{Range = u\sqrt{\frac{2h}{g}}}$
$v_{\mbox{just before touching ground}} = \sqrt{u^2+2gh}$
Title: Re: Projectile motion formulas
Post by: TrueTears on May 12, 2009, 07:15:38 pm: Draw a right angled triangle and use Pythagoras
Title: Re: Projectile motion formulas
Post by: naved_s9994 on May 12, 2009, 07:30:47 pm: Thanks TT

One day I have to pay you back some how....like SERIOUSLY !!

Thanks once again !!
Title: Re: Projectile motion formulas
Post by: naved_s9994 on May 12, 2009, 07:42:42 pm: For full para....max height...you got sin squared...

how do you put that into calc...DOESNT WORK??

Quote from: TrueTears on April 20, 2009, 02:21:36 pm
Ok these are the complete formulas:

For FULL parabolic path:

(' $u$ ' stands for initial speed not the horizontal or vertical.)

$\mbox{Total time of flight}$ $= \frac{2usin\theta}{g}$
$\mbox{Height_{max}}$ $= \frac{u^2sin^2 \theta}{2g}$
$\mbox{Range}$ $= \frac{u^2sin(2 \theta)}{g}$

For HALF parabolic path:

( $'u'$ stands for horizontal speed)

$\mbox{Total time of flight = \sqrt{\frac{2h}{g}}}$
$\mbox{V{vertical speed} = \sqrt{2gh}}$
$\mbox{Range = u\sqrt{\frac{2h}{g}}}$
$v_{\mbox{just before touching ground}} = \sqrt{u^2+2gh}$
Title: Re: Projectile motion formulas
Post by: GerrySly on May 12, 2009, 07:49:00 pm: $(\sin{(x)})^2$
Title: Re: Projectile motion formulas
Post by: TrueTears on May 12, 2009, 07:50:57 pm: $sin^2(x) = sin(x)^2$
Title: Re: Projectile motion formulas
Post by: naved_s9994 on May 12, 2009, 07:51:48 pm: tnx
Title: Re: Projectile motion formulas
Post by: naved_s9994 on May 12, 2009, 08:02:36 pm: Last QUESTION on Projectiless.....Umm. When is the angle BELOW the horizontal and when is it ABOVE the horizontal?
Title: Re: Projectile motion formulas
Post by: TrueTears on May 12, 2009, 08:09:24 pm: What do you mean by below or above? Normally the question would just say "...makes a $35^o$ angle with the horizontal"
Title: Re: Projectile motion formulas
Post by: naved_s9994 on May 12, 2009, 08:15:28 pm: ok then