An eagle is flying horizontally at 6.8 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?

Answer 1

(a) To find the time it takes for the fish's speed to double, we can use the formula for the distance fallen under constant acceleration:

v = v0 + at

where v is the final velocity, v0 is the initial velocity (0 m/s in this case, since the fish is only moving horizontally before being dropped), a is the acceleration due to gravity (approximated as 9.8 m/s^2), and t is the time we want to solve for.

We want to find the time it takes for the fish's speed to double, so we can set v = 2 * 0 m/s = 0 m/s. Plugging in the values, we get:

0 m/s = 0 m/s + (9.8 m/s^2) * t

Solving for t, we find that the time it takes for the fish's speed to double is:

t = (0 m/s) / (9.8 m/s^2) = 0 s

(b) Now we want to find how much additional time is required for the speed to double again. This means we want to find the time it takes for the fish's speed to go from 2 * 0 m/s = 0 m/s to 4 * 0 m/s = 0 m/s.

Using the same equation as before, we can set v = 4 * 0 m/s = 0 m/s and then solve for the time:

0 m/s = 0 m/s + (9.8 m/s^2) * t

Again, solving for t, we find that the time it takes for the fish's speed to go from 2 * 0 m/s = 0 m/s to 4 * 0 m/s = 0 m/s is:

t = (0 m/s) / (9.8 m/s^2) = 0 s

So the additional time required for the speed to double again is also 0 seconds.