An eagle is flying horizontally at 7.9 m/s with a fish in its claws. It accidentally drops the fish. (a) How much time passes before the fish's speed doubles? (b) How much additional time would be required for the speed to double again?
To solve this problem, we need to understand the concept of acceleration due to gravity and apply the equations of motion.
Let's start by calculating the time it takes for the fish's speed to double.
(a) How much time passes before the fish's speed doubles?
1. We know that when an object falls freely under gravity, it experiences a constant acceleration of 9.8 m/s^2. The initial velocity of the fish is horizontal, so this acceleration won't affect its horizontal motion.
2. Since the fish is dropped, it starts with an initial velocity of 0 m/s in the vertical direction.
3. We need to find the time it takes for the fish's vertical velocity to increase from 0 m/s to double its initial velocity.
4. The equation we can use to find the time is:
v = u + at
where:
v = final velocity (2 * initial velocity)
u = initial velocity (0 m/s)
a = acceleration due to gravity (9.8 m/s^2)
t = time
5. Rearranging the equation to solve for time, we have:
t = (v - u) / a
Substituting values into the equation, we get:
t = (2 * 0 - 0) / 9.8
t = 0 s
Therefore, the time required for the fish's speed to double is 0 seconds. This means that immediately after being dropped, the fish will start falling downward with a speed that is double its initial horizontal velocity.
(b) How much additional time would be required for the speed to double again?
Since the fish already had a vertical speed double its initial horizontal velocity, no additional time is required for the speed to double again. The fish would continue to fall freely with a constant acceleration of 9.8 m/s^2, doubling its speed after every time interval of 0 seconds.
In summary:
(a) The time it takes for the fish's speed to double is 0 seconds.
(b) No additional time is required for the speed to double again. It would instantly double each time it is measured.