How to Determine the Initial Velocity for Simultaneous Impact?

In summary: So the final equation becomesv=u+at^2where "a" is the acceleration. So in summary, the thrown rock has an initial velocity of 1, and the dropped rock has an initial velocity of 0. The dropped rock hits the ground at the same time as the thrown rock because the initial velocity of the dropped rock is zero.
  • #1
Joza
139
0

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 hight, 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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  • #2
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?
 
  • #3
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?
 
  • #4
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?
 
  • #5
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.
 
  • #6
Ah, do you know the equations of motion for constant acceleration?
 
  • #7
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!
 
  • #8
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?
 
  • #9
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
 

FAQ: How to Determine the Initial Velocity for Simultaneous Impact?

What is constant acceleration?

Constant acceleration is the rate of change of velocity that remains the same over a period of time. It is represented by the symbol "a" and is measured in units of meters per second squared (m/s^2).

How do you calculate the constant acceleration?

The constant acceleration can be calculated by dividing the change in velocity (Δv) by the change in time (Δt). This can be represented by the formula a = Δv/Δt.

What is the difference between constant acceleration and uniform motion?

Constant acceleration is when the velocity of an object changes at a constant rate, while uniform motion is when the velocity remains constant. In other words, in constant acceleration, the object is speeding up or slowing down, while in uniform motion, the object is moving at a constant speed.

How can you solve a constant acceleration problem?

To solve a constant acceleration problem, you can use the formula v = u + at, where v is the final velocity, u is the initial velocity, a is the constant acceleration, and t is the time. You can also use the formula x = ut + 1/2at^2, where x is the distance traveled.

What are some real-life examples of constant acceleration?

Some real-life examples of constant acceleration include a car accelerating from a stop, a skydiver falling towards the ground, and a rocket taking off into space. In all of these cases, the velocity of the object is changing at a constant rate over time.

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