17 replies to this topic

[mimbo193]

Ok so my teacher hasnt taught us about free falls yet, but he expects us to do the homework. I guess its not much different then regular kinematics equations, but I am still severely confused.

there's a few problems i would really appreciate help with, that i have attempted. But I'll only put one for now.

(neglect air resistance)

You throw a stone vertically upwards with an initial speed of 6.0m/s from a third-story office window. If the window is 12m above the ground, find (a) the time the stone is in flight and (b) the speed of the stone just before it hits the ground.

So far i have gotten a value for A using the formula, y=Vi(t) + 1/2gt^2

i got 12=6.00t+4.9t^2. which i used to get a quadratic. from the quadratic formula (a=4.9, b=6, c=-12) i got that t=2.30s (because time cant be negative). but i don't know if this is correct, and i also don't know how to solve the second part of the equation.

Please help:)

[Sweeney]

It's basically a matter of plugging all the values into the equations of motion in different ways.
It's not particularly difficult.

Working out the speed at impact is the easiest bit, as long as you remember that the stone's velocity at the peak of the throw will be zero.
Edit: You will, of course, have to work out the height of the peak first.

Edit two: The problem with your initial equation is that you have the initial velocity going in the wrong direction. It should be negative with respect to the displacement and acceleration, because the throw is upwards; the opposite direction to the displacement/gravity.

[mimbo193]

to be honest, i have no idea how to calculate the height of the peak.

but if i were to use 2gy=Vf^2-Vi^2, would Vf=15.33m/s?
also did i solve the time correctly?(sorry i just noticed your edit) i redid quadratic using -6 for B and got that t=1.06s

Edited by mimbo193, 20 September 2010 - 01:28 PM.

[Sweeney]

to be honest, i have no idea how to calculate the height of the peak.

but if i were to use 2gy=Vf^2-Vi^2, would Vf=15.33m/s?
also did i solve the time correctly?(sorry i just noticed your edit) i redid quadratic using -6 for B and got that t=1.06s

By my calculation, that final velocity is too low.
I haven't calculated the time, but that seems way too low. I shall take a look.

[mimbo193]

By my calculation, that final velocity is too low.
I haven't calculated the time, but that seems way too low. I shall take a look.

to get this I used
g=9.8
y=12
Vi=0
and Vf was the unknown

[Sweeney]

to get this I used
g=9.8
y=12
Vi=0
and Vf was the unknown

Ah, there's your problem. The initial velocity is 6m/s.
(Technically negative, but in this case it actually doesn't matter)

I checked your total time, and I got a result closer to your first than your second. I'm not entirely sure what you're doing differently, though.

[mimbo193]

i had a feeling i did that wrong... i wasnt sure because you said "Working out the speed at impact is the easiest bit, as long as you remember that the stone's velocity at the peak of the throw will be zero." i assumed that meant that the initial speed would be 0 seeing as when it reaches the peak its speed changes.

but ok i shall recalculate using 6 as the initial velocity.

2(9.8((12)=235.2
235.2=Vf^2-36

271.2=Vf^2--use sq rt to find Vf

16.5m/s/s=Vf?

also should i type out my work for time? i kind of want to know what im doing wrong >.<

Edited by mimbo193, 20 September 2010 - 01:43 PM.

[Sweeney]

i had a feeling i did that wrong... i wasnt sure because you said "Working out the speed at impact is the easiest bit, as long as you remember that the stone's velocity at the peak of the throw will be zero." i assumed that meant that the initial speed would be 0 seeing as when it reaches the peak its speed changes.

but ok i shall recalculate using 6 as the initial velocity.

Yeah, but if you calculate from the peak, the displacement would be bigger ^^
It's easier to do it from the window. I was wrong when I spoke earlier; looking at the wrong equation.

also should i type out my work for time? i kind of want to know what im doing wrong >.<

Please do

[mimbo193]

y=VI(t) + 1/2 G t^2

12=-6.00t + 4.9t^2
0=4.9t^2 -6.00t -12

B+/- sq rt (b2+4ac
_____________
2ac
A=4.9
B=-6
C=-12

-6+16.47/9.8=1.07 (sorry i forgot to round up)
-6-16.47/9.8=-2.29

Edited by mimbo193, 20 September 2010 - 01:53 PM.

[Turtleboy]

I didn't read through all the posts you guys made but I did see you didn't get the problem yet.

Maybe it'll be easier if you split it into two parts.

Part 1: From the third story office window to when it's velocity is 0 (i.e the peak of the flight)
-you can calculate the distance traveled (d)

Part 2: Take that distance you just calculated, tack on the distance from the window to the ground (its traveling form the peak to the ground now) and calculate the velocity.
Keep in mind what Sweeney said about the velocity at the peak of the throw.

[Warlord.]

Can't you use:

v = v_i + at

To solve how long it take to get from it's initial velocity of 6.0m/s to it's peak, substitute 0m/s for v.

0 = 6 + -9.8 * t
t = (0 - 6)/-9.8

t = 0.6122448979591837s

Then to find how high that peak is, use the position formula with the time value we just found:

x = x_i + v_it + 1/2 * at²

In our case:
x_i = 12m
v_i = 6m/s
a = -9.8m/s²
t = 0.6122448979591837s

Therefore, it's peak height:

x = 12 + 6 * 0.6122448979591837 + 1/2 * -9.8 * 0.6122448979591837²
x = 13.83673469387755m

Next, you could just set up a freefall from the peak height and find the positional velocity at 0m.

I haven't done physics in a whileee so I'm not sure if this is all right, but these should be some steps.

Edited by Warlord., 20 September 2010 - 06:38 PM.

[Turtleboy]

Eek, forgot about calculating the time traveled.

When you do that you have to do it for each section.and then add up the time.

You have your two equations:

(vf)^2 = (vi)^2 + 2ad
d = (vi)t + (1/2)a(t^2)

Edit:
Warlock, I think Sweeney was trying not to give the answer but just help guide him through the steps.
This way he can experiment with what equations he can and cannot use and not just have an answer

Edited by Turtleboy, 20 September 2010 - 01:59 PM.

[Sweeney]

y=VI(t) + 1/2 G t^2

12=-6.00t + 4.9t^2
0=4.9t^2 -6.00t -12

B+/- sq rt (b2+4ac
_____________
2ac
A=4.9
B=-6
C=-12

-6+16.47/9.8=1.07 (sorry i forgot to round up)
-6-16.47/9.8=-2.29

Ah, it's (-B +/- sqrt(b2+4ac))/2a)
There's your problem.

That will give you the correct time of 2.29s, and I believe you're already set to get the correct final velocity.

There's no need to divide the problem into two parts.

[mimbo193]

Eek, forgot about calculating the time traveled.

When you do that you have to do it for each section.and then add up the time.

You have your two equations:

(vf)^2 = (vi)^2 + 2ad
d = (vi)t + (1/2)a(t^2)

Edit:
Warlock, I think Sweeney was trying not to give the answer but just help guide him through the steps.
This way he can experiment with what equations he can and cannot use and not just have an answer

Her* sorry im a chick lol

plus it does make me feel better when i finally reach the answer lol

Ah, it's (-B +/- sqrt(b2+4ac))/2a)
There's your problem.

That will give you the correct time of 2.29s, and I believe you're already set to get the correct final velocity.

There's no need to divide the problem into two parts.

wow i feel really dumb now xD lol its always the little negative signs that i forget that mess me up!

Ok thank you for the help it is very appreciated

i still have 2 more... that i need help on if you dont mind. And just so everyone knows i did successfully do 6 other problems on my own... these are just the ones that I do not understand (because the answers are not in the back to guide me & they are lengthy and intimidating).

The worlds tallest building is 509 m tall (1667ft), and has 101 stories. The outdoor observation deck is on the 89th floor, and the two high speed elevators that service it reach a peak of 1008m/min on the way up and 610m/min on the way down. Assuming these peak speeds are reached at the midpoint of the run and that the accelerations are constant for each leg of the runs, (a) what are the accelerations for the up and down runs? (b) how much longer is the trip down than the trip up?

Im thinking for B that you need the height of the 89th floor, but im not sure how to find that.
and for A i need some guidance as well

Edited by mimbo193, 20 September 2010 - 03:03 PM.

[Sweeney]

Working out the height of any particular level is easy. Just work out the height per story using the two values given.
Not sure why you're after the 87th level though.

[darkmagegurl]

(-B +/- sqrt(b2+4ac))/2a)

no it's not. it's [-b+ sqrt (b^2 - 4ac)] / 2a

[mimbo193]

Working out the height of any particular level is easy. Just work out the height per story using the two values given.
Not sure why you're after the 87th level though.

i hit the wrong key.... sorry i was reading and typing at the same time. And im stressed because of my homework load.

Thats what i thought but i didnt want to make a false assumption.
Using 509m as the base and dividing it by 101, you get that each story is 5.03m correct?

then from this you divide 447.67 by the trip times (1008m/min and 610m/min)?



I think that was just a typo because i subtracted there and got the same values that Sweeney did

Edited by mimbo193, 20 September 2010 - 02:50 PM.

[Sweeney]

no it's not. it's [-b+ sqrt (b^2 - 4ac)] / 2a

My bad.
It's still plus or minus after the negative b, but it should be minus 4ac, you're right.