May 26, 2020, 10:55:35 pm

### AuthorTopic: Projectile Motion  (Read 2930 times) Tweet Share

0 Members and 1 Guest are viewing this topic.

#### iamdan08

• Victorian
• Posts: 698
• VCE Survivor
• Respect: +7
##### Projectile Motion
« on: March 25, 2008, 06:33:56 pm »
0
I was looking at some questions from the "Jacaranda Physics 2" textbook (i use heinmenn) and i came across this question:

A gymnast wants to jump a distance of 2.5m, leaving the ground at an angle of 28 degrees. With what speed must the gymnist take off?

Any help would be great
2007-08 VCE - Accounting, Texts & Traditions, Methods, Chem, Physics, Lit

2011 Bachelor of Biomedicine (Completed) @ The University of Melbourne
2012 Doctor of Medicine (Second Year) @ The University of Melbourne

#### ed_saifa

• Victorian
• Posts: 911
• Respect: +5
##### Re: Projectile Motion
« Reply #1 on: March 25, 2008, 07:30:35 pm »
0
i use the same book! the questions are really ambiguous and this one is no different.
[IMG]http://img411.imageshack.us/img411/2506/avatarcg3.png[/img]
(\ /)
(0.o)
(><)
/_|_\

"It's not a community effort"
"It's not allowed. Only death is a valid excuse"
"Probably for the first time time this year I was totally flabbergasted by some of the 'absolute junk' I had to correct .... I was going to use 'crap' but that was too kind a word"
"How can you expect to do well when
-you draw a lemon as having two half-cells connected with a salt bridge
-your lemons come with Cu2+ ions built in" - Dwyer
"Why'd you score so bad?!" - Zotos

#### iamdan08

• Victorian
• Posts: 698
• VCE Survivor
• Respect: +7
##### Re: Projectile Motion
« Reply #2 on: March 25, 2008, 07:35:43 pm »
0
lol...i agree. I have tried approaching the questions a few different ways. Can't seem to get it though!
2007-08 VCE - Accounting, Texts & Traditions, Methods, Chem, Physics, Lit

2011 Bachelor of Biomedicine (Completed) @ The University of Melbourne
2012 Doctor of Medicine (Second Year) @ The University of Melbourne

#### unknown id

• Victorian
• Trendsetter
• Posts: 131
• Respect: +1
##### Re: Projectile Motion
« Reply #3 on: March 25, 2008, 07:36:00 pm »
0
To answer this question, its better to use only the first half of the parabolic path taken by the gymnast. Hence, it's best to start off by listing all your known values:

uh = ucos(28) m/s ---> can be derived from a vector diagram
uv = usin(28) m/s

ah = 0 m/s^2
av = -10 m/s^2

vh = ucos(28) m/s ---> stays the same
vv = 0 m/s

xh = 1.25 m

Note: xh, vh and vv are the values taken at the end of half of the parabolic path.

Now, we can use the constant acceleration formulae to calculate the time taken by the gymnast to get to the maximum turning point of her path. The following calculation makes use of horizontal projectile values:

$x = \frac{1}{2}(u + v)t$

$1.25 = \frac{1}{2}(2u \cdot cos(28))t$

$t = \frac{1.25}{ucos(28)}$

Now that we've got a value for time, we can substitute this value into another formula. The following calculation makes use of vertical projectile values:

$v = u + at$

$0 = usin(28) - 10 \left(\frac{1.25}{ucos(28)} \right)$

$10(1.25) = (usin(28))(ucos(28))$

$12.5 = u^{2}sin(28)cos(28)$

$u^{2} = \frac{12.5}{sin(28)cos(28)}$

$u = 5.5 m/s$
« Last Edit: March 25, 2008, 07:44:46 pm by unknown id »
VCE Outline:
2007:   Accounting [48]

2008:   English [44], Maths Methods [50], Specialist Maths [41], Chemistry [50], Physics [44]

ENTER: 99.70

#### iamdan08

• Victorian
• Posts: 698
• VCE Survivor
• Respect: +7
##### Re: Projectile Motion
« Reply #4 on: March 25, 2008, 07:52:33 pm »
0
Awsome! Thats the right answer.    How do we know that its the horizontal speed that needs to be found and not the speed at the angle of 28 degrees? Do we just assume?
2007-08 VCE - Accounting, Texts & Traditions, Methods, Chem, Physics, Lit

2011 Bachelor of Biomedicine (Completed) @ The University of Melbourne
2012 Doctor of Medicine (Second Year) @ The University of Melbourne

#### unknown id

• Victorian
• Trendsetter
• Posts: 131
• Respect: +1
##### Re: Projectile Motion
« Reply #5 on: March 25, 2008, 07:58:52 pm »
0
I found the speed at the angle of 28 degrees, which was the asked for value.

Remember:

uh = $ucos(28)$

"u" was the value I calculated for and not "uh"
VCE Outline:
2007:   Accounting [48]

2008:   English [44], Maths Methods [50], Specialist Maths [41], Chemistry [50], Physics [44]

ENTER: 99.70

#### iamdan08

• Victorian
• Posts: 698
• VCE Survivor
• Respect: +7
##### Re: Projectile Motion
« Reply #6 on: March 25, 2008, 08:00:26 pm »
0
Oh...i see! Understand now. Thanks again.
2007-08 VCE - Accounting, Texts & Traditions, Methods, Chem, Physics, Lit

2011 Bachelor of Biomedicine (Completed) @ The University of Melbourne
2012 Doctor of Medicine (Second Year) @ The University of Melbourne

#### sxcalexc

• Victorian
• Forum Obsessive
• Posts: 330
• Respect: +1
##### Re: Projectile Motion
« Reply #7 on: March 28, 2008, 05:27:42 pm »
0
Nice solution unknown id but there is a much easier way - our old friend, the range formula

$R = \frac{v{^2} \times \sin 2x}{g}$

$2.5 = \frac{v{^2} \times \sin 2(28)}{10}$

$25 = v{^2} \times \sin 2(28)$

$v^2 = \frac{25}{sin 2(28)}$

$v = 5.5 m/s$

#### cara.mel

• Guest
##### Re: Projectile Motion
« Reply #8 on: March 28, 2008, 05:56:08 pm »
0
The range formula is not your friend. VCAA don't like it.

It's especially not your friend when the range they calculate on the paper they worked out using g=9.8, and so if you used that value when working out v or whatever we needed to find you got it wrong (2007 unit 3)

Anyway, be wary of using it, they'd *much* prefer it if you split it into horizontal and vertical components.
But by all means double check with it

#### ed_saifa

• Victorian
• Posts: 911
• Respect: +5
##### Re: Projectile Motion
« Reply #9 on: March 28, 2008, 05:58:48 pm »
0
This "range" formula looks intriguing.
[IMG]http://img411.imageshack.us/img411/2506/avatarcg3.png[/img]
(\ /)
(0.o)
(><)
/_|_\

"It's not a community effort"
"It's not allowed. Only death is a valid excuse"
"Probably for the first time time this year I was totally flabbergasted by some of the 'absolute junk' I had to correct .... I was going to use 'crap' but that was too kind a word"
"How can you expect to do well when
-you draw a lemon as having two half-cells connected with a salt bridge
-your lemons come with Cu2+ ions built in" - Dwyer
"Why'd you score so bad?!" - Zotos

#### sxcalexc

• Victorian
• Forum Obsessive
• Posts: 330
• Respect: +1
##### Re: Projectile Motion
« Reply #10 on: March 28, 2008, 07:45:07 pm »
0
The range formula is not your friend. VCAA don't like it.

It's especially not your friend when the range they calculate on the paper they worked out using g=9.8, and so if you used that value when working out v or whatever we needed to find you got it wrong (2007 unit 3)

Anyway, be wary of using it, they'd *much* prefer it if you split it into horizontal and vertical components.
But by all means double check with it

Ah.. right didn't know that. Thanks for the advice! As for the g value of 9.8 if you see the previously posted solution it has to assume a value of 9.8 or 10 also (the acceleration value).

#### cara.mel

• Guest
##### Re: Projectile Motion
« Reply #11 on: March 28, 2008, 07:48:33 pm »
0
Sorry, my post wasn't very clear, ignore me.

On the 2007 unit 3 exam they had a question where you had to find something. If you used the range value they gave you, you got it wrong because the question writer calculated the range using a=9.8 instead of a=10, but if you used whatever you were 'expected' to the long way you got it right.

Range is also bad because people tend to forget you have to land at the same height, you lot would be clever enough to realise that but not everyone else is.

#### sxcalexc

• Victorian
• Forum Obsessive
• Posts: 330
• Respect: +1
##### Re: Projectile Motion
« Reply #12 on: March 28, 2008, 07:50:53 pm »
0
Ah, I can see that  Damn saved so much time lol. I'll have to take this up with my teacher, see what he thinks.

#### Collin Li

• VCE Tutor
• Victorian
• ATAR Notes Legend
• Posts: 4966
• Respect: +17
##### Re: Projectile Motion
« Reply #13 on: March 28, 2008, 09:14:50 pm »
0
All those prepackaged formulae rely on particular assumptions. You are free to use them as long as you get the answer right, but if you get the answer wrong, you can expect 0 out of 3 or 4 marks (as they tend to give for questions like these).

It is better to work it out, and use a prepackaged formula on your cheatsheet to check your answer. You should derive them yourself, so you are aware of the assumptions. The typical assumption used in most (not all - some are generalised) prepackaged formulae is that your starting height is the same as your landing height.