Kinematics problem two stones falling one with velocity one at rest

AI Thread Summary
A kinematics problem involves two stones: one dropped from rest and another thrown downward at 11 m/s from the same height. Both stones land simultaneously, prompting a discussion about the equations of motion to determine the cliff's height. The user initially struggles with the relevant formulas but receives guidance on setting up equations for velocity and position. Ultimately, it is clarified that the second stone was thrown one second later, leading to a calculated height of approximately 128.6 meters for the cliff. The discussion emphasizes the importance of understanding motion equations in solving gravity-related problems.
bark00
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hi please can sum one help me,

a stone is dropped from he top of a cliiff at rest another stone it thrown downwards from the same point at 11ms-1. The two stones land at the same time, what is the height of the cliff,:frown:

how do u work this out i have no idea,:cry:

thankyou in advance

:smile: bark00:smile:
 
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Which equations do you think you're supposed to use?
 
thankyou for helping me,

I am not sure,
ive tried using combinations of v=u+at and s=1/2 (u+v)t but i don't know where to begin.
 
bark00 said:
thankyou for helping me,
im not sure,
ive tried using combinations of v=u+at and s=1/2 (u+v)t but i don't know where to begin.

This is a gravity problem...Can you set up the equations for both velocity and position in the horizontal (x) and vertical (y) direction ?


ok, before actually giving specific information, i suggst you read this tutorial

The second formula you gave is not known to me. Are you sure of it ?
 
Are you sure this is the question?
So you release one stone from rest and you throw the other one vertically down at 11 m/s from the same cliff and they land omn the ground at the same time? How's that possible?

Either I misunderstood the question, you copied it wrong or the height is zero and there would be no cliff.
 
hi thankyou all for the help i think i go it now i got the awnser as 128.6 metres approx
 
sorry missed out second one was trown a second later.
 
marlon said:
The second formula you gave is not known to me. Are you sure of it ?

The second formula is equivalent to s = V_{avg}t for constant acceleration.
 
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