NationStates Jolt Archive

thanks for all of your help guys

new problem on pg 3

would the time be equal to .4183?

please let me know, this is the last time i could give an answer, and if its wrong i get no credit at all despite my hard work

would the time be equal to .4183?

please let me know, this is the last time i could give an answer, and if its wrong i get no credit at all despite my hard work

This is all for credit, not for practice?

May I ask if your school has an academic honesty policy?

Do you think its fair that you'll be graded on your ability + the answers you get here, whereas other students will be graded just on their ability?

I'm not trying to be rude, but I've been both a teacher (not in physics) and a student (in physics), and what you're doing seems like it might infringe on the right of your classmates to be assessed fairly, and it doesn't really help you to learn the material.

would the time be equal to .4183?
please let me know, this is the last time i could give an answer, and if its wrong i get no credit at all despite my hard work
Well you know that the answer to the time question can't be zero ;)....let's see, rate of descent is 4.9m per second squared right?
So then could you model the descending speed with y = 20 - 4.9x^2 ?

The time to reach the highest point is:

t=v/a=20/9.8=2.041 seconds

v - velocity

a - acceleration

The time needed to reach the bottom of the hole:

t*t=distance/a=30/9.8=3.061=>t=1.75

Total time the rock stays in the air:

2.041+1.75=3.791 seconds

As far as I can remember, time in this instance is directly related to the velocity of the object and the distance travelled. So:

t = delta v/distance.

But you only have the distance travelled in the hole (10m) not the distance the object goes in the air, so you can't work out the time taken. I think so correct me if I'm wrong, been a few years since I did one of these

thanks for the input, and heres is how i got that answer for the velocity

vi= 20m/s
t=?
a=9.8m/s^2
vf=?
d=-10


i did the vf^2=vi^2+2ad equation

and got: 24.41, so i just switched the sign to -24.4

The time to reach the highest point is:

t=v/a=20/9.8=2.041 seconds

v - velocity

a - acceleration

The time needed to reach the bottom of the hole:

t*t=distance/a=30/9.8=3.061=>t=1.75

Total time the rock stays in the air:

2.041+1.75=3.791 seconds


thanks for the try, but thats wrong. that was my first answer

But you only have the distance travelled in the hole (10m) not the distance the object goes in the air, so you can't work out the time taken.

The time can be worked out, I believe its about 2.49 seconds (edit: for the time from apogee to ground), but I could be wrong.

EDIT: Sorry, forgot to add back in the time up, 2.04. Now I'm getting 4.53

If you start at 20 m/s up, divide by 9.8 to give you the number of seconds before it reaches the apogee, mutiply that by (20-0)/2 (the average velocity during time in flight up) to get its heigth at apogee where gravity has overcome upward initial velocity. Its then at that height (20.4 meters), and begins to fall with acceleration -9.8 m/s^2. You can use d = (1/2)(a)(t^2),
set d at 30.4 meters (20.4 height plus the 10 meter hole), and solve for t.

Use the same formula as before:

y(t) = 0,5 a t^2 + v(o) t + y(0)

Here:
a = g (which is about -9,8 m/s^2; but use your textbook value)
v(0) = 20
y(0) = 0
y(final) = -10

Put the values in and solve for t.

Completely taking air resistance out of the equation, the rock will have the same velocity is started with at the point of release. From the point of release, it will accelerate for an additional 10m at a rate of 9.81 m/s^2 (gravity)

First claculate how far (and time) the rock will take to stop moving when going up. When you have the total distance and time going up, you can simply add 10m to the distance and calculate the velocity at the bottom using g.

If I give you any info besides that it wouldn't be fair. Also, always balance your units. Make sure the equation you use gives the units you want. It can be easy to screw up if you aren't paying attention to what you are actually calculating.

Use the same formula as before:

y(t) = 0,5 a t^2 + v(o) t + y(0)

Here:
a = g (which is about -9,8 m/s^2; but use your textbook value)
v(0) = 20
y(0) = 0
y(final) = -10

Put the values in and solve for t.

would he need to solve for t at y(t) = -10, to account for the hole?
EDIT: Sorry, my fault, you counted for it

The time can be worked out, I believe its about 2.49 seconds, but I could be wrong.

If you start at 20 m/s up, divide by 9.8 to give you the number of seconds before it reaches the apogee, mutiply that by (20-0)/2 (the average velocity during time in flight up) to get its heigth at apogee where gravity has overcome upward initial velocity. Its then at that height (20.4 meters), and begins to fall with acceleration -9.8 m/s^2. You can use d = (1/2)(a)(t^2),
set d at 30.4 meters (20.4 height plus the 10 meter hole), and solve for t.

so that would give me 4.5312 seconds right?

thanks for the try, but thats wrong. that was my first answer
Yes, I don't know why I assumed the rock would reach a height of 20 meters. It's 40.82 meters in fact (I hope I'm right now :p).
d=a*t*t=9.8*2.041*2.041=40.82

The total distance the rock falls is 40.82+10=50.82 meters, so t=2.28 seconds.

Total time in the air: 4.32 seconds.

Do you know your equations of motion?

s = ut +1/2at^2
v = u + at
etc

and what is gravity assumed to be? -10 or -9.8?

would he need to solve for t at y(t) = -10, to account for the hole?

Correct. The answer is slightly below 5; but he must do that himself :P

The velocity of the rock on the bottom of the hole would be:

v=a*t=9.8*4.32=42.36m/s

I hope I was right this time. :)

now im gettin 4.75 as the total time it was in the air including the time it fell down the hole

thanks everyone, but i think i got it. i think its 4.75s and now im gonna input it. thanks again, couldnt have done it without you

now im gettin 4.75 as the total time it was in the air including the time it fell down the hole

You were warmer before :P (unless I made a calculating error)

You were warmer before :P (unless I made a calculating error)

oh no, but are you accounting for the while time its in the air or just the time it took to fall down the hole? i calculated it as taking.67s to fall down the hole and 4.08 to get there

You were warmer before :P (unless I made a calculating error)

I'm getting 4.53 (being 2.04 up and 2.49 back down and through the hole)

oh no, but are you accounting for the while time its in the air or just the time it took to fall down the hole? i calculated it as taking.67s to fall down the hole and 4.98 to get there

I do not calculate those seperately; I just set y(t)=-10 and solve the equation for t.

I'm getting 4.53 (being 2.04 up and 2.49 back down and through the hole)

nvm i got the same answer as you, must have had a calculating error in my last post

heres how i did it:

vf=vi + a(t)
-24.41=-20 + 9.8(t)
t= .45s (to fall downt the hole)

The answers are 24.4 m/s and 4.53 seconds.

s = ut + 1/2at^2
v = u + at
v^2 = u^2 + 2as

Use them wisely, my friend.

The rock peaks at 20.4 meters (vf^2=vi^2+2ad) and it takes 2.04 seconds (vf=vi+at) Add 10 on to that, resulting in 30.4m for the net fall on the other side and solve for Vf in the equation:

2d=(vf^2-vi^2)/(2a)

60.8=(vf^2-0)/(19.6)
1191.68=vf^2

vf=34.52m/s at the bottom of the pit. To find T, divide 34.52 by 9.8 to get 3.52 seconds for the time to hit the bottom. Oh, and the net time would be 5.56 seconds.

thanks guys, youre all awesome

:fluffle:

I'm getting 4.53 (being 2.04 up and 2.49 back down and through the hole)

Same final answer - didn't bother to calculate the up and down.

here are the final, correct answers

vf: -24.41
t: 4.53

The rock peaks at 20.4 meters (vf=vi+at) Add 10 on to that, resulting in 30.4m for the net fall and solve for Vf in the equation:

2d=(vf^2-vi^2)/(2a)

60.8=(vf^2-0)/(19.6)
1191.68=vf^2

vf=34.52m/s at the bottom of the pit.

I'm getting 24.4 m/s

(calculating from apogee, -9.8 = a with d =(20.4 +10) = 30.4

d= (1/2)(a)(t^2), set d at 30.4, solve for t, t= 2.49

falling for 2.49 seconds at a= 9.8 gives impact velocity of 24.4 m/s in hole

I'm getting 24.4 m/s

(calculating from apogee, -9.8 = a with d =(20.4 +10) = 30.4

d= (1/2)(a)(t^2), set d at 30.4, solve for t, t= 2.49

falling for 2.49 seconds at a= 9.8 gives impact velocity of 24.4 m/s in hole

Yes, I see what I did now. I used 30.4 in the wrong place by accident. That's why it's about 10 off of your answer.

The trick is to break the problem up into little pieces.

Throwing the rock upwards

v = u + at

t = (v - u) / a = (0 - 20) / -9.8 = 2.04 seconds. This is time to maximum height.

Now we use this in s = ut + 1/2at^2.

s = (20*2.04) + 1/2*(-9.8)*(2.04^2) = 20.41m. This is maximum height.

Thus on the return trip, the object falls s = 30.41m.

Rock falling

Again we use s = ut + 1/2at^2.

30.41 = 0t + 1/2*(9.8)*(t^2) <--- Note acceleration is now considered positive, as it acts in the direction of motion!

t = sqrt(30.41 / 4.9) = 2.49 seconds. This is time to fall from maximum height.

Thus total time in air is 2.04 + 2.49 = 4.53 seconds.

Wait until you get into trajectories and components. On one high-school exam, I was able to calculate the clearance between a tennis ball and the net, with the trajectory not being symmetrical about the net. ;)

Still remember the buzz as I scored 88% in that exam. :D

The trick is to break the problem up into little pieces.

I tend to gravitate toward that approach because it answers lots of questions about the same problem, and gives multiple chances to check for reasonability.

i hate how some of the parts to these questions are very easy to calculate, but then the others are mad hard

http://i26.photobucket.com/albums/c144/slicknick727/vectors.jpg

a)What is the angle alpha between vectors E and F in the figure?
b)Use components to determine the magnitude of G=E+F.
c)Find theta, the angle between G and the x-axis.

A)71.6 degrees
B)3
C) ?

im stumped on c, can someone explain how to find this angle out?
i know that the angle between e and the x-axis is 45degrees, and that the angle from F to the y-axis is 26.6degrees

Still remember the buzz as I scored 88% in that exam. :D

Imagine how well you'd have done if you weren't buzzed.

c) looks like 90 degrees.

A useful mental trick to adding vectors, is mentally moving one vector (say, E) such that it begins where the other vector ends. Vector G then looks like a vertical line along the y-axis with magnitude 3.

I tend to gravitate toward that approach because it answers lots of questions about the same problem, and gives multiple chances to check for reasonability.

True - but to get the time and final velocity it is not needed (you can get the final velocity by inserting the t your found in dy(t)/dt ).

For the actual understanding of what is happening it may be better though - just a tad bit more work.

True - but to get the time and final velocity it is not needed (you can get the final velocity by inserting the t your found in dy(t)/dt ).

For the actual understanding of what is happening it may be better though - just a tad bit more work.

You're right, its not necessary for the two part question, and on an exam where time is a factor, it'd be nice to scope one expression.

For applications, though, part of me would rather have a bunch of solved aspects.

For the actual understanding of what is happening it may be better though - just a tad bit more work.
This is where true knowledge lies, though. A lot of my classes at school and university, were about fulfilling the requirements of the syllabus, and in industry, that does not get you very far. And it made it all boring and unsatisfying. You feel so much more confidence about those areas where you really can take the problem apart. I regret I did not understand more.

You should probably be doing these by yourself - youre not going to learn by asking other people for answers. Check textbooks and online physics help pages before giving up and asking here.

thanks, i actually suspected it to be around 90 when i drew it, but i wasnt sure if it was exactly 90