Are two objects in free fall hitting the water at the same time?

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Homework Help Overview

The problem involves two stones being thrown vertically downward from a height of 50.0 meters, with the first stone thrown with an initial speed of 2.00 m/s and the second stone thrown 1.00 second later. The inquiry focuses on the timing of their impacts with the water below and the conditions under which they can hit the water simultaneously.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the timing of the stones' impacts and the implications of the second stone being thrown 1 second after the first. There is confusion regarding the use of "t-1" in calculations and whether this means the second stone hits the ground before the first.

Discussion Status

Participants are exploring the relationship between the timing of the throws and the conditions for simultaneous impacts. Some guidance has been offered regarding the mechanics of the stones' motions, but questions remain about the interpretation of the timing and the calculations involved.

Contextual Notes

There is an ongoing discussion about the assumptions regarding the initial velocities and the timing of the throws, as well as the implications of the stones hitting the water simultaneously despite being thrown at different times.

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Homework Statement



An inquisitive physics student and mountain climber climbs a 50.0m high cliff that overhangs a calm pool of water. He throws two stones vertically downward, 1.00 s apart, and observes that they cause a single splash. The first stone has an initial speed of 2.00 m/s. (a) How long after release of the first stone do the two stones hit the water? (b) What initial velocity must the second stone have if the two stones are to hit the water simultaneously? (c) What is the speed of each stone at the instant the two stones hit the water?

Homework Equations


X=Xo-0.5gt^2

The Attempt at a Solution


a) X=-2t-4.9t^2
50=-2t-4.9t^2
t=2.99s

(b) X=V(t-1)-4.9(t-1)^2...

The thing that I don't understand is that the question states that both stones hit the ground at the same time (splash together), but in solving this problem we use "t-1" which would mean that the second stone hits the ground one second before the first stone. Am I interpreting the question wrong. Could you please explain fully
 
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Ab17 said:

Homework Statement



An inquisitive physics student and mountain climber climbs a 50.0m high cliff that overhangs a calm pool of water. He throws two stones vertically downward, 1.00 s apart, and observes that they cause a single splash. The first stone has an initial speed of 2.00 m/s. (a) How long after release of the first stone do the two stones hit the water? (b) What initial velocity must the second stone have if the two stones are to hit the water simultaneously? (c) What is the speed of each stone at the instant the two stones hit the water?

Homework Equations


X=Xo-0.5gt^2

The Attempt at a Solution


a) X=-2t-4.9t^2
50=-2t-4.9t^2
t=2.99s

(b) X=V(t-1)-4.9(t-1)^2...

The thing that I don't understand is that the question states that both stones hit the ground at the same time (splash together), but in solving this problem we use "t-1" which would mean that the second stone hits the ground one second before the first stone. Am I interpreting the question wrong. Could you please explain fully
Welcome to the PF.

The 2nd stone is thrown 1 second after the first, but with a higher initial vertical velocity. That is how it can catch up to the first stone just as they reach the water. Does that help?
 
I understand what you are pointing out. But I am confused if both stones do touch the ground at the same time because the second stone take 1.99s to touch ground while the first takes 2.99s
 
Ab17 said:
I understand what you are pointing out. But I am confused if both stones do touch the ground at the same time because the second stone take 1.99s to touch ground while the first takes 2.99s
That matches them being thrown 1 second apart, no? :smile:
 
So the time at which they reach the ground is not equal?
 
Ab17 said:
So the time at which they reach the ground is not equal?
They reach the surface of the water at the same time Tfinal. They are thrown at two different starting times. Their flight distances are equal. That's how you solve for the actual flight time durations...
 
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Ok thanks a lot I think I understand a little bit. Really appreciate your help. God bless you
 
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