- #1

- 205

- 0

so i am trying to use the equation x1=xo+tvo+.5at^2

then x2=xo+(t+1.6)vo+.5a(t+1.6)^2

or should i set my zero point somewhere else like where stone 2 is when the distance between the stones is 36 m?

please give me a lead

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- Thread starter Rasine
- Start date

- #1

- 205

- 0

so i am trying to use the equation x1=xo+tvo+.5at^2

then x2=xo+(t+1.6)vo+.5a(t+1.6)^2

or should i set my zero point somewhere else like where stone 2 is when the distance between the stones is 36 m?

please give me a lead

- #2

hage567

Homework Helper

- 1,509

- 2

You are given the difference in height between them at a certain time, so you need to find the time t that they are 36m apart.

- #3

- 960

- 0

I usually try to find the simplest physically intuitive approach. The way then i look at this problem is to "ignore" acceleration snce its the same for both and therefore cancels once both are in free fall and gaining velocity at the same rate. This does not mean that the distance between them is constant. So off the top of my head, I'd first figure the velocity after 1.6 sec=at. this is 15.68m/s. To reach 36m would require the time at this speed. Then backtrack for position. I may be all wet, is so someone willbe sure to say so.

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