. An inquisitive physics student and mountain climber climbs a 50.0-m 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 was the initial velocity of the second stone? (c) What is the velocity of each stone at the instant the two hit the water?
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To solve this problem, we can use the kinematic equations of motion for an object undergoing free-fall.
(a) How long after release of the first stone do the two stones hit the water?
Since the first stone is released with an initial speed of 2.00 m/s, and both stones are thrown vertically downward, we can use the equation for displacement to find the time of flight.
The formula for displacement in the vertical direction is given by:
d = ut + (1/2)gt^2
Where:
d = displacement (50.0 m, since the cliff is 50.0 m high)
u = initial speed (2.00 m/s)
t = time
g = acceleration due to gravity (approximately 9.8 m/s^2)
Substituting the given values:
50.0 m = (2.00 m/s) * t + (1/2) * (9.8 m/s^2) * t^2
Rearranging the equation and setting it equal to zero:
(1/2) * (9.8 m/s^2) * t^2 + (2.00 m/s) * t - 50.0 m = 0
We can solve this quadratic equation for t using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
In this equation, a = (1/2) * (9.8 m/s^2), b = 2.00 m/s, and c = -50.0 m.
Solving for t, we get two values: t = 1.01 s and t = -9.90 s. Since time cannot be negative in this case, we discard the negative value. Therefore, the two stones hit the water 1.01 seconds after the first stone is released.
(b) What was the initial velocity of the second stone?
The second stone was released 1.00 s after the first stone. Since both stones are thrown vertically downward, the only difference between them is the time of release. Therefore, the initial velocity of the second stone is equal to the acceleration due to gravity (approximately 9.8 m/s^2) multiplied by the time delay of 1.00 s. Thus, the initial velocity of the second stone is 9.8 m/s.
(c) What is the velocity of each stone at the instant the two hit the water?
Since the stones are thrown vertically downward, we can use the following equation to find the velocity at any given time:
v = u + gt
Where:
v = final velocity
u = initial velocity
g = acceleration due to gravity
t = time
For the first stone:
u1 = 2.00 m/s
g = 9.8 m/s^2
t = 1.01 s (from part a)
Substituting these values into the equation:
v1 = 2.00 m/s + (9.8 m/s^2) * 1.01 s = 11.9 m/s
For the second stone:
u2 = 9.8 m/s (from part b)
g = 9.8 m/s^2
t = 1.01 s (from part a)
Substituting these values into the equation:
v2 = 9.8 m/s + (9.8 m/s^2) * 1.01 s = 19.8 m/s
Therefore, the velocity of the first stone at the instant it hits the water is 11.9 m/s, and the velocity of the second stone at the instant it hits the water is 19.8 m/s.
To solve this problem, we can use the equations of motion and the principles of free fall. Let's go through each part of the problem step by step:
(a) How long after release of the first stone do the two stones hit the water?
We know that the first stone has an initial speed of 2.00 m/s and it falls for a certain time until it hits the water. The acceleration due to gravity is acting downward, so we can use the equation of motion for vertical motion:
h = vt + 0.5gt^2
Where:
- h is the height of the cliff (50.0 m),
- v is the initial velocity (2.00 m/s),
- g is the acceleration due to gravity (-9.8 m/s^2), and
- t is the time it takes for the stone to reach the water.
Since the second stone is released 1.00 second after the first stone, we can subtract 1.00 second from the time it takes for the first stone to reach the water to find the time at which the second stone hits the water.
Let's set up the equation for the first stone:
50.0 = (2.00)t + 0.5(-9.8)t^2
Simplifying this equation gives us a quadratic equation. We can solve it to find the value of "t".
(b) What was the initial velocity of the second stone?
Since the two stones hit the water at the same time, the second stone falls for "t" seconds. We need to find the initial velocity of the second stone. We can use the equation of motion:
v = u + gt
Where:
- v is the final velocity (which is 0 at the instant the stone hits the water),
- u is the initial velocity of the second stone, and
- g is the acceleration due to gravity (-9.8 m/s^2).
By substituting the known values into the equation and solving for "u", we can find the initial velocity of the second stone.
(c) What is the velocity of each stone at the instant they hit the water?
At the instant the first stone and the second stone hit the water, their velocities are both 0 m/s. This is because they're at rest after falling from the cliff.
By solving the equations in parts (a) and (b), we can find the answers to each question. Let's solve the equations and calculate the values for (a), (b), and (c).
a. Vo*t + 0.5g*t^2 = d.
2t + 4.9t^2 = 50.
4.9t^2 + 2t - 50 = 0.
Use Quadratic Formula and get:
t = 2.99681.
b. Vo*3 + 4.9*3^2 = 50.
Vo*3 + 44.1 = 50.
3Vo = 50 - 44.1 = 5.9.
Vo = 5.9 / 3 = 1.97 m/s.
c. Vf^2 = Vo^2 + 2g*d.
Vf^2 = 2^2 + 19.6*50 = 984.
Vf = 31.4 m/s. = Final velocity.