Answer :
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.
=======================================================
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.
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.
=======================================================
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.
Answer:
a) First stone, Vf₁ = 31.38 m/s
b) Second stone, Vf₂ = 34.81 m/s
Explanation:
1) Data:
First stone:
Vo₁ = + 2 m/s (positive sign is understood as throw in the same direction of the acceleration, i.e. downward).
h = 50 m
t₁ = t
Vf₁ = ?
Second stone:
t ₂ = t - 1
Vf₂ = ?
2) Solution
i) First stone, final velocity:
Vf² = Vo² + 2gh
Vf² = (2 m/s)² + 2(9.81m/s²)(50m) = 984 m²/s²
Vf = 31.38 m/s
ii) First stone time, t
Vf = Vo + gt ⇒ t = [Vf - Vo] / g = [31.38 m/s - 2m/s] / 9.81 m/s² = 3.0 s
iii) Second stone time, t - 1
t₂ = t - 1 = 3.0s - 1.0s = 2.0s
iv) Second stone initial velocity, Vo₂
h = Vo₂ t + gt₂² / 2 ⇒ 50m = Vo₂ (2.0s) + 9.81m/s² (2.0s)² / 2
Vo₂ = [ 50m - 19.62m] / (2.0s) = 15.19 m/s
v) Second stone final velocity, Vf₂
Vf₂ = Vo₂ + gt₂ = 15.19 m/s + 9.81 m/s² (2.0s) = 34.81 m/s
