Projectile Motion Problem

[laijiawen]

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.

[kamil9876]

What did I do wrong?

In order to help you with that, it would be good if you showed us what you did.

[laijiawen]

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
vertical velocity v=u+at
                        =10x0.94
                        =9.4m/s
v=square root (9.4^2+5.32^2)
  =10.8m/s

[bomb]

Edit: should have looked at your working -.-

[laijiawen]

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.

[kamil9876]

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.

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

[kevvy]

projectile motion is such a bitch.

[aznboy50]

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 into the formula because we usually ignore air resistance and any other friction..

[laijiawen]

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.

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

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?

[aznboy50]

There is no force acting on the glass in the upwards direction, how can it possibly up?

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.

[kamil9876]

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 into the formula because we usually ignore air resistance and any other friction..

why did you set v=0? The horizontal motion is constant hence (which gives you the equation we had before)

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.

[laijiawen]

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 into the formula because we usually ignore air resistance and any other friction..

why did you set v=0? The horizontal motion is constant hence (which gives you the equation we had before)

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.

okay I got it now. . Now I need help with this one,
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.

[bomb]

Posting answer in a sec..

[aznboy50]

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

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.

[bomb]

Ahh shit, I remember now, give me a sec It can be done I think.