#### ~christina~

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**A climber throws 2 stones 1.00s apart and they hit at same time h=50ft**

**height of cliff = 50ft**I think there is enough info now...

**1. Homework Statement**

A climber climbs a cliff overhanging a calm pool of water and throws 2 stones vertically down 1.00 s apart. The climber then observes that they cause a single splash. The first stone has a initial v of 2.00m/s

a) how long will it take after the release of the stones do the 2 stones hit the water.

b) what is the inital velcocity of 2nd stone if they are to hit simultaneously?

c) What is the speed of each at the instant the 2 stones hit the water?

**2. Homework Equations**

If I'm not incorrect in which eqzn I used...

then Position as a function of velocity and time...

xf= xi + 1/2(vxi + vxf)t

**3. The Attempt at a Solution**

**a.)**

d= 50 m ===>since it is starting from 0 if I'm not incorrect and flung downward then the final distance is 50 so it's negative 50...I think

diference in time thrown is = 1.00s

vi stone 1 = 2.00m/s

a= -9.80m/s^2

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xf= xi + vxi (t) + 1/2 (ax)(t^2)

-50m= 2.00t + .5(-9.80m/s^2)

0= 50m+ 2.00t -4.9t^2

through quadradic formula I found that

t=

__-2.00+/-rad (2.00)^2 - 4*(-4.9)*50__

-9.8

and for that I found that

**t= 3.40s**since the neg can't be used as a valid time

_____________________

Then for b)

I equated the 2 equations leaving out vxi for one side and for the time since they say a difference of 1.00s I went and added it to the time found for the first stone and used that but I'm not sure about that. (Assuming that the sencond stone was thrown after the first one) Is acceleration -9.8? since the person is throwing the stone not dropping it I wasn't sure about that.

xi-xf + vxi*(t)+ 1/2*ax*t^2 = xi-xf + vxi*(t)+ 1/2*ax*t^2

stone 1-------------------------stone 2

-50 + 2.00m/s*3.40s + 1/2*-9.80m/s^2*(3.40s)^2= -50 + vxi*4.40s+ 1/2*(-9.80m/s^2)*(4.40s)^2

-99.84= -144.864 + vxi*4.40s

45.024 = vxi*4.40s

vxi= 10m/s

_____________________________

For

**C.)**I don't know how to approach that..which equation do I use?

I know it's a kinematic one but I'm not sure which to use....

Thanks

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