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Rtael
Sep 8th, 2009, 05:30 PM
Just in an introductory physics class and having trouble working this out...

A stone is dropped from the roof of a building; 1.30 http://session.masteringphysics.com/render?units=s after that, a second stone is thrown straight down with an initial speed of 18.0 http://session.masteringphysics.com/render?units=m%2Fs, and the two stones land at the same time.

I have to figure out:
How long did it take the first stone to reach the ground?
How high is the building?
What are the speeds of the two stones just before they hit the ground?

I attempted using the formula x = x(0) + v(0)t + (1/2)at^2
and got 0 = x(0) + 0 + (1/2)(-9.8)(t^2)
and 0 = x(0) + 18t + (1/2)(-9.8)(t-1.3)^2
and worked that out to t = 1.6 seconds
unfortunately that was wrong, lol
any help?

Rtael
Sep 8th, 2009, 05:31 PM
those units came out weird but it's 1.3 s and 18.0 m/s

Sally Struthers
Sep 8th, 2009, 05:43 PM
it has been a long time so if this is wrong don't blame me..

:lol:

how high is hte building?

2as =(v1)^2 - (v0)^2

2(-9.8m/s^2)s = 0 - (18 m/s )^2

s=(- 324m^2/s^2)/(-19.6 m/s^2)

s= 16.53m

Rtael
Sep 8th, 2009, 07:49 PM
nope, not the right answer :p

Sally Struthers
Sep 8th, 2009, 08:31 PM
well what is the right answer?

Sep 8th, 2009, 08:34 PM
1
a = 9.8
v = 0
u = 18

v = u + at
0 = 18 + 9.8t
-18 = 9.8t
t = 1.83s (+1.30)
t = 3.13s

2
s = ut + 1/2at^2
s = 18x1.83 + 1/2x9.8x1.83^2
s = 32.94 + 16.41
s = 49.35m

3
Stone 1
v = u + at
v = 0 + 9.8 x 3.13
v = 30.67ms^-1

Stone 2
v = u + at
v = 18 + 9.8x1.83
v = 35.93ms^-1

probably completely wrong, but i tried :angel:

Rtael
Sep 8th, 2009, 08:36 PM
Also wrong. And I don't know what the right answer is...I'm doing it online and i have a certain number of tries to get it right.

Sep 8th, 2009, 08:49 PM
Also wrong. And I don't know what the right answer is...I'm doing it online and i have a certain number of tries to get it right.

masteringphysics sucks...our school tried it with us for a couple weeks, was rubbish though so haven't used it since...

i feel sorry for you :)

AleOrtu
Sep 8th, 2009, 09:03 PM
Remove the "(easy)" word in the title of the thrad :scratch:

FrOzon
Sep 8th, 2009, 09:23 PM
How long did it take the first stone to reach the ground? 2,88sec
How high is the building? 40.7m
What are the speeds of the two stones just before they hit the ground?
Stone 1 = 28.3 m/s
Stone 2 = 33.5 m/s

Sally Struthers
Sep 8th, 2009, 09:23 PM
ok ..

y =y0 +v0t -1/2gt^2

y = 0 + 18(1.3) -1/2(9.8)(1.3)^2

y = 15.119 meters

question a

y = y0 +vot - 1/2gt^2

15.119 m = 0+0 -1/2(9.8)t^2

t = 1.76 seconds

question c
stone 1
v= v0 -gt
v = 0-9.8(1.76)
v = 17.248 m/s

stone 2

v= 18 -9.8(1.76+1.3)
v = 11.99 m/s

Apoorv
Sep 8th, 2009, 09:34 PM
s = (0)*t + 0.5*9.8*t*t = 18*(t-1.3) + 0.5*9.8*(t-1.3)*(t-1.3)
=> 4.9t*t = 18t - 23.4 + 4.9t*t - 12.74t + 8.28
=> 5.26t = 14.72
=> t = 2.8 sec. (approx.)

height, h = 0.5*9.8*2.8*2.8 = 38.416 m

speed of stone 1,
v = 0 + 9.8 * 2.8 = 27.44 m/s

Speed of stone 2,
v = 18 + 9.8 * 1.5 = 32.7 m/s

Is this solution correct, I used to study physics about 8 years back. So it may be wrong. I just took g as positive.
because it is in the direction of throw of stone i.e downwards.

Sally Struthers
Sep 8th, 2009, 09:35 PM
:lol: all different answers. no one knows what they are doing :haha:

FrOzon
Sep 8th, 2009, 09:49 PM
s = (0)*t + 0.5*9.8*t*t = 18*(t-1.3) + 0.5*9.8*(t-1.3)*(t-1.3)
=> 4.9t*t = 18t - 23.4 + 4.9t*t - 12.74t + 8.28
=> 5.26t = 14.72
=> t = 2.8 sec. (approx.)

height, h = 0.5*9.8*2.8*2.8 = 38.416 m

speed of stone 1,
v = 0 + 9.8 * 2.8 = 27.44 m/s

Speed of stone 2,
v = 18 + 9.8 * 1.5 = 32.7 m/s

Is this solution correct, I used to study physics about 8 years back. So it may be wrong. I just took g as positive.
because it is in the direction of throw of stone i.e downwards.

Almost correct! ;) Just a small error in line 3: 23.4 - 8.28 = 15.12, then you get the same numbers as I do :p
(Also a physicist :angel:)

Apoorv
Sep 8th, 2009, 10:12 PM
Almost correct! ;) Just a small error in line 3: 23.4 - 8.28 = 15.12, then you get the same numbers as I do :p
(Also a physicist :angel:)

Thanks!!:) I had a feeling I am making a mistake in calculation but I was too lazy to check :o

moby
Sep 8th, 2009, 11:03 PM
I used s = ut + 1/2(a)(t)^2. (Let starting position be 0.)

Taking downward direction as +ve,
write out s_1 (u_1 = 0, a_1 = g, t_1 = t) and s_2 (u_2 = 18, a_2 = g, t_2 = t - 1.3) and equate them. You get a linear equation; the t^2 terms cancel out.

Solving for t, we get:
t = [(18)(1.3) - 1/2(1.3)(1.3)g]/[18 - 1.3g]

If you use g = 9.8 ms^-2, you get 2.8743 s (5 s.f.)
If you use g = 9.80665 ms^-2, you get 2.8780 s (5 s.f.)

The rest of the question follows easily.

Rtael
Sep 9th, 2009, 12:05 AM
Okay, thanks guys! I was actually doing it (pretty) correct, but I missed changing the v(0)t to t-1.3 in the second equation instead of just t! Thanks to everyone who tried to help but especially frozon for the correct answers and to apoorv for writing out the equations and helping me see where I went wrong.

Rtael
Sep 9th, 2009, 12:06 AM
and moby though you were later :p

barboza
Sep 9th, 2009, 12:17 AM
Which college are u in?