How to Determine the Initial Velocity for Simultaneous Impact?

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Homework Help Overview

The problem involves two rocks: one is thrown straight up from a cliff's edge, and the other is dropped from the same edge after a certain time. The goal is to determine the initial velocity of the thrown rock so that both rocks hit the ground simultaneously. The cliff height, the time difference between the two rocks, and the acceleration due to gravity are provided.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the relationship between the time it takes for the dropped rock to hit the ground and the initial velocity of the thrown rock. There is exploration of the equations of motion and how they apply to the scenario, particularly regarding the total time the thrown rock is in the air.

Discussion Status

Participants are actively engaging with the problem, confirming each other's understanding of the time intervals involved and the equations of motion. Some express confusion about the upward motion of the thrown rock and its implications for the calculations. There is a recognition of the need to clarify the role of displacement in the equations being discussed.

Contextual Notes

Participants are working within the constraints of the problem as posed, including the assumption that the height of the cliff and the acceleration due to gravity are known. There is an acknowledgment of the complexity introduced by the upward motion of the first rock.

Joza
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Homework Statement



I have 2 rocks. One is thrown straight up from a cliff's edge. A certain amount of time later another rock is dropped from the cliffs edge.

I need to calculate the initial velocity of the rock 1 so that they hit the ground simultaneously. I have the cliff height, the time after rock 1 when rock 2 was thrown, and acceleration of course. I just need some pointers as I can't seem to get my thinking behind it.

Cheers!
 
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The only thing that is going to change is the initial velocity of the thrown rock. Which is what we're looking for. Would you agree that the dropped rock with always hit the ground in the same time interval since the height initial velocity and acceleration of gravity won't change?

Does this help at all?
 
Yes I agree with that. I have the time it takes to hit the ground from s=ut+(1/2)at^2.

The first rock is in the air for this time plus the time interval when the 2nd rock started. Correct? Is this needed?
 
Joza said:
Yes I agree with that. I have the time it takes to hit the ground from s=ut+(1/2)at^2.

The first rock is in the air for this time plus the time interval when the 2nd rock started. Correct?

Exactly correct! That's how much time the thrown rock will be in the air total. Do know how to go about solving from here?
 
Mmm...I'm not exactly sure yet. What confuses me is that it is going up and decelerating. Then it accelerates down again a distance the same as rock 2 plus and unknown distance.
 
Ah, do you know the equations of motion for constant acceleration?
 
v=u + at

s=ut + (1/2)at^2

v^2=u^2+2as


Are they relevant here? Maybe this is all really easy, it's probably so easy but my brain is tired!
 
Sure we don't the height in flight but s=ut + (1/2)at^2 doesn't require it does it? You know the initial position the rock was thrown and you know where it lands correct?

"s" is the final position of the rock, if we call the cliff 0 then the final position, the ground, is a negative amount of units below the cliff right?
 
Right...?
 
  • #10
Ok, I got it!


I realized that distance above the cliff is negligible because it's displacement is 0 back at the cliff when it comes down...
 
  • #11
Joza said:
Ok, I got it!


I realized that distance above the cliff is negligible because it's displacement is 0 back at the cliff when it comes down...

Yup! Nice
 

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