Post by: laijiawen on January 04, 2011, 12:03:03 am
Post by: kamil9876 on January 04, 2011, 12:13:41 am
Quote
What did I do wrong?
In order to help you with that, it would be good if you showed us what you did.
Post by: laijiawen on January 04, 2011, 12:24:43 am
1.1=5t^2
t=0.47s x 2
=0.94s
horizontal velocity Uh=5/0.94
=5.32m/s
vertical velocity v=u+at
=10x0.94
=9.4m/s
v=square root (9.4^2+5.32^2)
=10.8m/s
Post by: bomb on January 04, 2011, 12:29:45 am
Post by: laijiawen on January 04, 2011, 12:32:39 am
If it's Heinemann I wouldn't worry about it, they have ridiculous rounding.
Also, some textbooks use gravity as 9.8m/s.
I don't think its heinemann, the question is from a sheet that my teacher gave us to complete for the holidays, I did the whole sheet using a=10 so I dont think a=9.8.
Post by: kamil9876 on January 04, 2011, 01:06:17 am
Quote
Okay, x=vt+0.5at^2
1.1=5t^2
t=0.47s x 2
=0.94s
horizontal velocity Uh=5/0.94
=5.32m/s
Good. Right idea.
Quote
vertical velocity v=u+at
=10x0.94
=9.4m/s
No, we don't need any "vertical velocity". Read the question carefully, we want the speed of the car which is the initial horizontal velocity of the glass(also, how can a car have vertical velocity?). According to your calculation above this is 5.32. The mistake you made was that t=0.47 (no idea where the x 2 came from). This gives you:
horizontal velocity=5/0.47=10.638
Post by: kevvy on January 04, 2011, 01:24:19 am
Post by: aznboy50 on January 04, 2011, 09:11:32 am
A car left the road and collided with a tree. Glass from the windscreen was projected forward and landed in the grass 5.0m away. The average height of the windscreen was 1.1m above the ground. Determine the speed of the car at the time of impact. The answer says 10.65m/s while I got 10.8m/s. What did I do wrong? I tried alternating my calculations by taking into account sig figs and exact values, but I still got 10.75-10.8m. THANKS.
OK , first we need to find how long it takes for the windscreen hit the ground
Just a quick question Kamil, is the final horizontal velocity zero??
I would think the final velocity is not zero, as that implies that there is deceleration throughout the motion.
Why wouldn't this work in the forumula:
Also, I think you can put
Post by: laijiawen on January 04, 2011, 10:02:52 am
oh, i thought you had to times t by 2 cause the glass goes up and down. how do we know if the glass is half projectile or a part projectile where v=o when glass is at highest point?QuoteOkay, x=vt+0.5at^2
1.1=5t^2
t=0.47s x 2
=0.94s
horizontal velocity Uh=5/0.94
=5.32m/s
Good. Right idea.Quotevertical velocity v=u+at
=10x0.94
=9.4m/s
No, we don't need any "vertical velocity". Read the question carefully, we want the speed of the car which is the initial horizontal velocity of the glass(also, how can a car have vertical velocity?). According to your calculation above this is 5.32. The mistake you made was that t=0.47 (no idea where the x 2 came from). This gives you:
horizontal velocity=5/0.47=10.638
Post by: aznboy50 on January 04, 2011, 11:58:45 am
You are simply looking at the glass, as though it is 1.1 metres above the ground. Gravity is acting upon it. No other forces are specified. Therefore, the glass will obviously go to the ground.
Post by: kamil9876 on January 04, 2011, 11:59:50 am
I would think the final velocity is not zero, as that implies that there is deceleration throughout the motion.
Why wouldn't this work in the forumula:
Also, I think you can putinto the formula because we usually ignore air resistance and any other friction..
why did you set v=0? The horizontal motion is constant hence
Quote
oh, i thought you had to times t by 2 cause the glass goes up and down. how do we know if the glass is half projectile or a part projectile where v=o when glass is at highest point?
You don't times it by 2. Remember the glass starts from the top and just drops (while also moving horizontally). I'm presuming that you caught on to the habit of multiplying by 2 by doing problems that involve a particle going up and then down? The glass here just does the second half of such a journey. The initial vertical velocity is 0 as you correctly subbed in in your first post.
Post by: laijiawen on January 04, 2011, 02:29:41 pm
okay I got it now. :D. Now I need help with this one,
I would think the final velocity is not zero, as that implies that there is deceleration throughout the motion.
Why wouldn't this work in the forumula:
Also, I think you can put
why did you set v=0? The horizontal motion is constant hence(which gives you the equation we had before)
Quoteoh, i thought you had to times t by 2 cause the glass goes up and down. how do we know if the glass is half projectile or a part projectile where v=o when glass is at highest point?
You don't times it by 2. Remember the glass starts from the top and just drops (while also moving horizontally). I'm presuming that you caught on to the habit of multiplying by 2 by doing problems that involve a particle going up and then down? The glass here just does the second half of such a journey. The initial vertical velocity is 0 as you correctly subbed in in your first post.
A car has run into a fire hydrant and come to an abrupt stop. A suitcase tied to a rack on top of the car has been thrown off and landed on the roadside 11.6m away from the hydrant. The suitcase is found to have been 1.2m above the ground when it was still on the rack. Determine the impact speed of the car if the launching angle for the suitcase was 10degrees.
Post by: bomb on January 04, 2011, 04:54:53 pm
Post by: aznboy50 on January 04, 2011, 06:25:56 pm
Here's the picture, I'll try and latex and answer in a sec :) First time latexing, bear with me.
Time taken for the suitcase to hit the ground:
So, lets find the velocity
Someone that's actually good at physics please check this. I suck at physics, but wanted to help :P
This is incorrect.
The suitcase goes up and then down. You must make two separate equations.
I don't think we can do this question. We need to be given the mass of the bag, then we can solve using kinetic and gravitational potential energy.
Post by: bomb on January 04, 2011, 07:20:16 pm
Post by: aznboy50 on January 04, 2011, 07:22:20 pm
Ahh shit, I remember now, give me a sec :P It can be done.
Please do it!!!!!!!
I've been trying to solve this for over half an hour!!!
Post by: bomb on January 04, 2011, 07:35:13 pm
Post by: aznboy50 on January 04, 2011, 07:37:46 pm
I mean this question :
A car has run into a fire hydrant and come to an abrupt stop. A suitcase tied to a rack on top of the car has been thrown off and landed on the roadside 11.6m away from the hydrant. The suitcase is found to have been 1.2m above the ground when it was still on the rack. Determine the impact speed of the car if the launching angle for the suitcase was 10degrees.
I'm telling you now, you can't solve it.
Post by: bomb on January 04, 2011, 07:39:31 pm
Edit, found the question I did, it was very similar BUT it had the time at max height.
Post by: aznboy50 on January 04, 2011, 07:46:44 pm
I know which question you meant, just wanted to see if it's the same one I did a while go. I remember doing a similar one. I do think there's information missing...
Edit, found the question I did, it was very similar BUT it had the time at max height.
That explains it...
Post by: laijiawen on January 04, 2011, 09:02:38 pm
Here's the picture, I'll try and latex and answer in a sec :) First time latexing, bear with me.The answer says 14.5m/s. I think you forgot to take into account the angle and stuff. This question stumps me ~.~. Plus it has a Challenging written next to it by the teacher. ;D
Time taken for the suitcase to hit the ground:
So, lets find the velocity
Someone that's actually good at physics please check this. I suck at physics, but wanted to help :P
Post by: kamil9876 on January 04, 2011, 10:35:33 pm
edit: just two points:
(1) your teacher didn't mention whether the angle is up or down.
(2) doesn't seem very realistic that there should be a "launching angle" since by newton's (first?) law it seems as though there would need to be some force applied at such an angle in order to change the direction of the suitcase.
But with some (possibly bullshit) assumptions, it is solvable with just kinematics.
Post by: s2penguin on January 04, 2011, 10:37:44 pm
You definitely don't need the mass. It's solvable using just kinematics.
Could you please do it, I too cannot do it?
Post by: evaever on January 04, 2011, 10:53:14 pm
horizontal: u=v, s=11.6, t? s=ut
eliminate t to find v=14.5
Post by: s2penguin on January 05, 2011, 08:54:30 am
Post by: evaever on January 05, 2011, 12:07:41 pm
It has to be up, otherwise it goes through the roof of the car.
(2) doesn't seem very realistic that there should be a "launching angle" since by newton's (first?) law it seems as though there would need to be some force applied at such an angle in order to change the direction of the suitcase.
It is realistic because when the car hits the obstacle the rear may tilt upwards and project the suitcase at an angle, keep in mind the car is not a point.
Post by: kamil9876 on January 05, 2011, 03:32:42 pm
Post by: evaever on January 05, 2011, 03:43:59 pm
when a horizontal jet of water hits a wall, water goes in all directions after reflection, not just horizontally
Post by: laijiawen on January 05, 2011, 07:19:07 pm
vertical: u=vtan10, a=-10, s=-1.2, t? s=ut+0.5at2so is this the answer?
horizontal: u=v, s=11.6, t? s=ut
eliminate t to find v=14.5
Post by: bomb on January 05, 2011, 07:23:05 pm
Post by: laijiawen on January 05, 2011, 07:30:59 pm
Post by: bomb on January 05, 2011, 07:35:41 pm
surely someone could solve this??? I really need to solve this, so I can have my work completed as I am a perfectionist and need everything done before I move on. =[
You're a perfectionist..for now :P So was I, it lasts like two weeks.
Post by: evaever on January 05, 2011, 07:53:28 pm
vertical: u=vtan10, a=-10, s=-1.2, t? s=ut+0.5at2so is this the answer?
horizontal: u=v, s=11.6, t? s=ut
eliminate t to find v=14.5
definitely
Post by: evaever on January 05, 2011, 07:54:50 pm
No, because like me ha has put distance as 1.2m, when the suitcase goes up first...
-1.2 is the displacement, not the distance
Post by: evaever on January 05, 2011, 07:59:32 pm
How did you get the initial vertical velocity u= vtan10?
The horizontal component of the velocity vector is v, the vertical component of the velocity vector must be vtan10, so that the velocity vector makes a 10 deg angle. Draw a right triangle to see it.
Post by: kamil9876 on January 05, 2011, 08:06:54 pm