A mountain climber stands at the top of a 50-meter cliff hanging over a calm pool of water. The climber throws two stones vertically 1 second apart and observes that they cause a single splash when they hit the water. The first stone has an initial velocity of +2.0 meters per second. What will the velocity of each stone be at the instant both stones hit the water?

One of two things is true about this question: EITHER it can't happen as you've described it, OR you've left out some vital information.
-- IF the first stone was thrown downward with an initial speed and the second one was dropped from rest 1 second later, then the second one can never catch up with the first one, and they can never hit the water together.
-- IF the first stone was thrown downward with an initial speed, AND the second one was released 1 second later, AND they actually do hit the water together, THEN the second stone must have been given an initial downward speed greater than 2 m/s, otherwise it could never catch up with the first one.
Note: The masses and weights of the stones are irrelevant and not needed.
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An afterthought . . . . .
If the first stone was tossed UP at 2 m/s . . . that could be the meaning of the prominent plus-sign that you wrote next to the 2 . . . then it rises for (2/9.8) second, then begins to fall, and passes the mountain climber's hand on the way down (4/9.8) second after he tossed it, falling at the same 2.0 m/s downward.
From there, it still has 50m to go before it hits the water. 50 = 2 T + 1/2 G T² 4.9 T² + 2 T - 50 = 0 T = 3 seconds
The first stone hits the water 3 seconds after passing the mountain climber's hand on the way down at a downward speed of 2.0 m/s. In that 3 seconds, it gains (3 x 9.8) = 29.4 m/s of additional speed, hitting the water at (29.4 + 2) = 31.4 m/s .
This is all just a guess, assuming that the 2.0 m/s was an UPWARD launch. Maybe I'll come back later and calculate the second stone.

Answered by AL2006 | 2024-06-10

a) First stone, Vf₁ = 31.38 m/s
b) Second stone, Vf₂ = 34.81 m/s ;

Answered by Edufirst | 2024-06-11

The final velocity of the first stone when it hits the water is approximately 31.34 m/s, while the final velocity of the second stone is approximately 19.62 m/s. Both stones hit the water closely together despite being thrown at different times. This results in a combined splash upon impact.
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Answered by Edufirst | 2024-09-03