An eagle is flying horizontally at 5.8 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 free fall and acceleration due to gravity.
In this scenario, the fish is dropped from a stationary position in the eagle's claws and falls freely downward due to the force of gravity.
(a) To calculate the time it takes for the fish's speed to double, we can use the equation for free fall:
v = u + gt
where:
v = final velocity (which is twice the initial velocity in this case)
u = initial velocity (0 m/s since the fish was dropped)
g = acceleration due to gravity (9.8 m/s^2)
t = time
In this case, we want to find the time it takes for the fish's speed to double, so v = 2u = 2(0) = 0 m/s.
0 = 0 + (9.8)t
Simplifying the equation, we have:
9.8t = 0
Since t multiplied by any number will still be zero, we can say that the time it takes for the fish's speed to double is 0 seconds.
(b) Now, let's consider the additional time required for the speed to double again.
Given that the fish's initial speed is 0 m/s after being dropped, we need to calculate the time it takes for its speed to reach twice the initial speed once again. This time, the initial velocity is 0 m/s.
Again, we can use the equation for free fall:
v = u + gt
Let's denote the time it takes for the speed to double again as T.
2u = u + gT
Simplifying the equation, we have:
u = gT
Since the initial velocity is zero at the start of the second doubling, we can rewrite the equation as:
0 = gT
However, we know that the value of g is 9.8 m/s^2, so we can say that T multiplied by any number will still be zero.
Therefore, the additional time required for the fish's speed to double again is also 0 seconds.