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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 + 1/2at

^{2}

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 = D

_{down}- S

_{up}???

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.

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