A rock climber stands on top of a 56m -high cliff overhanging a pool of water. He throws two stones vertically downward 1.0s apart and observes that they cause a single splash. The initial speed of the first stone was 2.5(m/s)
How long after the release of the first stone does the second stone hit the water?
What was the initial speed of the second stone?
What is the speed of the first stone as it hits the water?
What is the speed of the second stone as it hits the water?
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To solve this problem, we can use the equations of motion under constant acceleration.
1. How long after the release of the first stone does the second stone hit the water?
Let's start by finding the time it takes for the first stone to hit the water. We can use the equation:
h = ut + (1/2)gt^2
Where:
h = height (56m)
u = initial velocity (2.5 m/s)
t = time taken for the first stone to hit the water
g = acceleration due to gravity (approximated as 9.8 m/s^2)
Rearranging the equation, we get:
t = (sqrt(2h/g) - u/g)
Substituting the given values:
t = (sqrt(2 * 56 / 9.8) - 2.5 / 9.8)
t = (sqrt(11.4286) - 0.2551)
t ≈ (3.38 - 0.2551)
t ≈ 3.12 seconds
Since the second stone is thrown 1.0s after the first stone, we can add 1.0s to get the total time taken for the second stone to hit the water:
Total time = t + 1.0s
Total time ≈ 3.12s + 1.0s
Total time ≈ 4.12 seconds
Therefore, the second stone hits the water approximately 4.12 seconds after the release of the first stone.
2. What was the initial speed of the second stone?
Since both stones are thrown vertically downward and only 1.0 second apart, the second stone will also have the same initial vertical velocity as the first stone, which is 2.5 m/s.
Therefore, the initial speed of the second stone is 2.5 m/s.
3. What is the speed of the first stone as it hits the water?
The speed of the first stone can be calculated using the equation:
v = u + gt
Where:
v = final velocity (speed of the first stone as it hits the water)
u = initial velocity of the first stone (2.5 m/s)
g = acceleration due to gravity (approximated as 9.8 m/s^2)
t = time taken for the first stone to hit the water (3.12 seconds)
Substituting the given values, we get:
v = 2.5 + 9.8 * 3.12
v ≈ 30.57 m/s
Therefore, the speed of the first stone as it hits the water is approximately 30.57 m/s.
4. What is the speed of the second stone as it hits the water?
Since the second stone has the same initial vertical velocity and falls for the same amount of time as the first stone, the speed of the second stone as it hits the water will also be approximately 30.57 m/s.
Therefore, the speed of the second stone as it hits the water is approximately 30.57 m/s.