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1D Kinematics - 2 stones thrown - where do they meet?

  • Thread starter Hemingway
  • Start date
  • #1
42
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Homework Statement



Two stones are thrown simultaneously, one straight upward from the base of a cliff and the other straight downward from the top of the cliff. The height of the cliff is 5.85 m. The stones are thrown with the same speed of 8.59 m/s. Find the location (above the base of the cliff) of the point where the stones cross paths.

answer = 2.36m --> not sure how to get there.

Homework Equations



s = ut + 1/2at2
perhaps some other 1D Kinematics equations to find other variables

The Attempt at a Solution



Assumptions:
1. SU (stone thrown upwards) velocity is upwards but acceleration is downwards (g)
2. SD (stone thrown downwards) velocity is down as is acceleration (g)
3. SD is faster than SU so they will not meet at the middle of height of cliff
4. (?) they will meet when SU reaches 0m/s velocity (disproven, i think)

other ideas:
5. they will not meet before the point of SU's 0m/s - at which point I don't know how to proceed
6. meeting will occur at t(d) = t(u)
7. Y = y(sd) - y(su)??


My last attempt:

hypothesis: they meet at SU zero point

v^2 = u^2 + 2as
s = v^2 - u^2/ 2a
s = 3.76m not the answer (2.36m)

so...

Stone downwards

v^2 = u^2 + 2as
= 8.59^2 + 2 (-9.81)(5.85)
v = -6.40m/s

t = (v-u)/a
= -6.40 - 8.59/-9.81
t = 1.53s

*******************

Stone upwards

y = 0m
a = 9.81m/s^2
u = 8.59m/s
y = ?

if y = 0

v = u + 2as
= 5.89 + 2(9.81)(0)
v = 5.89

does D = Ddown - Sup ???
if so what do I use as my displacement for stone upwards?

so.... I'm conceptually stuck. I don't have the best problem solving in physics..

Any hints regarding assumptions?

Many thanks,

H.
 
Last edited:

Answers and Replies

  • #2
56
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The best thing to do in such cases is to find the equations of motion for the each body, i.e. their exact location at any moment. Let the stones move along the y-axis (positive direction upwards).

For stone thrown upwards, you have

[tex]y_{up}=v_{0}t-\frac{1}{2}gt^{2},[/tex]

where [itex]v_{0}[/itex] is the initial velocity.

And for the stone thrown from the top of the cliff

[tex]y_{down}=h-v_{0}t-\frac{1}{2}gt^{2},[/tex]

where [itex]h[/itex] is the height of the cliff.

Using these equations and one more (condition for the stones to meet) it's easy to find the time and the sought coordinate.
 
  • #3
42
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Thank you very much! I'll try that out :)
 
  • #4
42
0
Thank you very much! I'll try that out :)
The best thing to do in such cases is to find the equations of motion for the each body, i.e. their exact location at any moment. Let the stones move along the y-axis (positive direction upwards).

For stone thrown upwards, you have

[tex]y_{up}=v_{0}t-\frac{1}{2}gt^{2},[/tex]

where [itex]v_{0}[/itex] is the initial velocity.

And for the stone thrown from the top of the cliff

[tex]y_{down}=h-v_{0}t-\frac{1}{2}gt^{2},[/tex]

where [itex]h[/itex] is the height of the cliff.

Using these equations and one more (condition for the stones to meet) it's easy to find the time and the sought coordinate.
Apologies, I am still lost. Ultimately does the answer = Yup - Ydown? I might not have the cerebral ability to work out this other condition..

Sorry :/
 
  • #5
42
0
Actually I worked it out don't worry!!! Thank you very much
 

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