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VCE Stuff => VCE Science => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Physics => Topic started by: kaanonball on January 20, 2009, 09:05:19 pm

Title: Projectile Motion Questions
Post by: kaanonball on January 20, 2009, 09:05:19 pm: 1.) A water skier at the Moomba Masters competition in Melbourne leaves a ramp at a speed of 50km/h and at an angle of 30 degrees. The edge of the ramp is 1.7m above the water. Calculate:

(a) the range of the jump [answer = 18m]
(b) the velocity at which the jumper hits the water. [answer = 15m/s at 37 degrees to the horizontal]

2.) A gymnast wants to jump a distance of 2.5m, leaving the ground at an angle 28 degrees. With what speed must the gymnast take off? [answer = 5.5m/s]

any help would be appreciated =]
Title: Re: Projectile Motion Questions
Post by: pHysiX on January 21, 2009, 11:16:54 am: Can't figure out question 1 yet but got question 2:

Let the velocity be x:
The vertical velocity would be xsin28 and the horizontal would be xcos28.

Now, we must find the total airborne time:
v=u+at
0= xsin28 - 10t
t= (1/10)xsin28

Hence, total airborne time is (1/5)xsin28 seconds

Now we know the range of the jump is 2.5m and it is constant horizontal velocity
Therefore:
d=vt
2.5=xcos28 * (1/5)xsin28
x^2 = 12.5/(cos28 * sin28)
x= 5.49m/s

Two s.f. ==> 5.5m/s

Sorry...i'm a bit n00b with latex =]
Title: Re: Projectile Motion Questions
Post by: bubble sunglasses on January 21, 2009, 01:40:38 pm: check out 6th and 7th replies
http://community.boredofstudies.org/421/physics/136267/projectile-motion-question.html

sorry, couldn't resist ;)
Title: Re: Projectile Motion Questions
Post by: pHysiX on January 21, 2009, 02:07:49 pm: ROFL HILARIOUS!
Title: Re: Projectile Motion Questions
Post by: IntoTheNewWorld on January 21, 2009, 03:49:55 pm: Dammit I keep getting 19m range for 1 a)

Sigh at physics
Title: Re: Projectile Motion Questions
Post by: kaanonball on January 21, 2009, 04:18:15 pm: Quote from: bubble sunglasses on January 21, 2009, 01:40:38 pm

check out 6th and 7th replies
http://community.boredofstudies.org/421/physics/136267/projectile-motion-question.html

sorry, couldn't resist ;)

LOL
Title: Re: Projectile Motion Questions
Post by: kurrymuncher on January 21, 2009, 04:41:19 pm: ROFL Enwiabe was so funny on BoS
Title: Re: Projectile Motion Questions
Post by: Mao on February 04, 2009, 10:14:24 pm: First question:

a)

initial horizontal velocity: $\frac{50}{3.6} \times \cos 30^o = 12.02\; ms^{-1}$

initial vertical velocity: $\frac{50}{3.6}\times \sin 30^o = 6.94\; ms^{-1}$
time to maximum: $\frac{0-6.94}{-10} = 0.69\; s$
vertical displacement: $ut + \frac{1}{2}at^2 = 6.94 \times 0.69 + \frac{1}{2}\times -10 \times 0.69^2 = 2.41\; m$
vertical distance (at maximum): $2.41 + 1.7 = 4.11\; m$

time to drop to zero: $ut + \frac{1}{2}at^2 = s \implies 0\times t + 5t^2 = 4.11\;\implies t=\sqrt{\frac{4.11}{5}} = 0.91\; s$

time (total): $0.69 + 0.91 = 1.60\; s$

displacement (horizontal): $1.60 \times 12.02 = 19.25 \; m$

b)
vertical velocity on landing: $0.91 \times 10 = 9.07\; ms^{-1}$

speed on landing: $\sqrt{9.07^2 + 12.02^2} = 15.06\; ms^{-1}$

angle to horizontal: $\sin^{-1}\frac{9.07}{15.06} = 37.03^o$

I think the answer given by the book is incorrect for part a)
Title: Re: Projectile Motion Questions
Post by: dejan91 on February 08, 2009, 11:17:45 pm: (http://i41.tinypic.com/2q3u25g.jpg)

LOL I had no idea how to put equations up on the forum...so my apologies if its unreadble or whatever.
Title: Re: Projectile Motion Questions
Post by: dejan91 on February 08, 2009, 11:23:37 pm: ^^Note that's the long way just to show HOW to get the answer. You could also use the range formula (which I'll post when I find).
Title: Re: Projectile Motion Questions
Post by: Mao on February 08, 2009, 11:50:22 pm: $R=v_h \times t = v_h \times 2 \times \frac{v_v}{g} = \frac{2\cdot v\cdot \cos \theta \cdot v \cdot \sin \theta}{g} = \frac{v^2\cdot \sin 2\theta}{g}$ (assuming the height of projection and landing are the same)
Title: Re: Projectile Motion Questions
Post by: dejan91 on February 09, 2009, 03:59:36 pm: Yup that's the one I was looking for. Much quicker way.