Author Topic: Daniel's Physics Questions (Read 9269 times) Tweet Share

danieltennis · « **Reply #15 on:** December 30, 2008, 03:30:14 pm »

Quote from: Flaming_Arrow on December 30, 2008, 03:28:19 pm

whats the answer?

6.5m

Flaming_Arrow · « **Reply #16 on:** December 30, 2008, 03:35:25 pm »

i did this but i got 6.4m

$mgh = \frac {m v^2}{2}$

$16 = 0.250 \cdot 10 \cdot h$

$h = 6.4m$

danieltennis · « **Reply #17 on:** December 30, 2008, 03:40:22 pm »

Quote from: Flaming_Arrow on December 30, 2008, 03:32:55 pm

i did this but i got 6.4m

$mgh = \frac {m v^2}{2}$

$16 = 0.250 \cdot 10 \cdot h$

$h = 6.4m$

Yeah, I also used that method to solve the question and got the same answer with you.
The problem is the question is in the "Projectile Motion" section not the conservation of momentum. So you have to use a projectile motion methodology to solve it, which I don't know how to.

NE2000 · « **Reply #18 on:** December 30, 2008, 04:31:29 pm »

$0.5mv^2 = KE_i =0.125(16)^2$

$KE_f = 16 J$

Hence

$GPE_f = 0.125(256) - 16 = 16 J$
$GPE_f = mgh = 16$
$0.25 \cdot 9.8 \cdot h = 16$
$h = 6.5 m$

danieltennis · « **Reply #19 on:** December 30, 2008, 04:34:02 pm »

Quote from: NE2000 on December 30, 2008, 04:31:29 pm

$0.5mv^2 = KE_i =0.125(16)^2$

$KE_f = 16 J$

Hence

$GPE_f = 0.125(256) - 16 = 16 J$
$GPE_f = mgh = 16$
$0.25 \cdot 9.8 \cdot h = 16$
$h = 6.5 m$

Thanks NE2000.

danieltennis · « **Reply #20 on:** January 01, 2009, 03:27:06 pm »

hey,

During training, an aerial skier takes off from a ramp that is inclined at 40 degrees to the horizontal and lands in a pool that is 10m below the end of the ramp. If she takes 1.5 seconds to reach the highest point of her trajectory, calculate:
A/ the speed she leaves the ramp

Thanks

NE2000 · « **Reply #21 on:** January 01, 2009, 04:41:32 pm »

Here goes

Consider vertical component of velocity

$u=V \sin 40$ where V = the speed she leaves the ramp

$a=-9.8, v=0, t= 1.5$

$v=u + at$

$0=V \sin 40 -14.7$

$14.7= V \sin 40$

$V = 22.9 m/s$

Is that right?

EDIT: used 9.8 for g instead of 10

danieltennis · « **Reply #22 on:** January 01, 2009, 04:52:28 pm »

hi, it's 22.9m/s.

NE2000 · « **Reply #23 on:** January 01, 2009, 05:02:58 pm »

Quote from: daniel. on January 01, 2009, 04:52:28 pm

hi, it's 22.9m/s.

lol, just use 9.8 for g instead of 10

/0 · « **Reply #24 on:** January 01, 2009, 05:21:50 pm »

VCE Physics uses $g=10ms^{-1}$
VCE Specialist Maths uses $g=9.8ms^{-1}$
(Checked exam papers to confirm)

Weird innit?

danieltennis · « **Reply #25 on:** January 01, 2009, 11:41:23 pm »

Quote from: daniel. on January 01, 2009, 03:27:06 pm

hey,

During training, an aerial skier takes off from a ramp that is inclined at 40 degrees to the horizontal and lands in a pool that is 10m below the end of the ramp. If she takes 1.5 seconds to reach the highest point of her trajectory, calculate:
A/ the speed she leaves the ramp

Thanks

b/ the time for which she is in mid-air?

vertical

$t=?, a=9.8ms^{-1}, x=21m, u=0$

$x=ut+1/2at^2$

$4.9t^2 = 21$

$t=2.07 seconds$

$Total time in air= 2.07 + 1.5$
$= 3.57 seconds$

Is this the fastest way in solving this problem?
Thanks

NE2000 · « **Reply #26 on:** January 02, 2009, 08:51:14 am »

Quote from: daniel. on January 01, 2009, 11:41:23 pm

Quote from: daniel. on January 01, 2009, 03:27:06 pm
hey,

During training, an aerial skier takes off from a ramp that is inclined at 40 degrees to the horizontal and lands in a pool that is 10m below the end of the ramp. If she takes 1.5 seconds to reach the highest point of her trajectory, calculate:
A/ the speed she leaves the ramp

Thanks

b/ the time for which she is in mid-air?

vertical

$t=?, a=9.8ms^{-1}, x=21m, u=0$

$x=ut+1/2at^2$

$4.9t^2 = 21$

$t=2.07 seconds$

$Total time in air= 2.07 + 1.5$
$= 3.57 seconds$

Is this the fastest way in solving this problem?
Thanks

That's how I would do it, if you go straight (i.e. don't separate the rising and falling part of the projectile motion) then you will get a quadratic I think. And it's best to avoid those.

danieltennis · « **Reply #27 on:** February 24, 2009, 07:39:59 pm »

Hey,

For those who have the Heinemann Physics textbook, I need help on q4 pg 56.

Any help will be appreciated.

thanks

TrueTears · « **Reply #28 on:** February 24, 2009, 07:45:27 pm »

what's the questions, I'll have a go XD

danieltennis · « **Reply #29 on:** February 24, 2009, 08:02:50 pm »

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