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

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Two stones are thrown from a 50.0m high cliff, with the first stone thrown downward at 2.00 m/s and the second thrown 1.00 second later. The first stone takes approximately 2.99 seconds to hit the water, while the second stone must have a higher initial velocity to reach the water simultaneously. The confusion arises from the timing of their throws; although they are thrown at different times, they can still splash together upon hitting the water. The second stone's flight time is shorter, allowing it to catch up to the first stone. Understanding the relationship between their throw times and velocities clarifies how they can hit the water simultaneously.
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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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