A pelican flying along a horizontal path drops
a fish from a height of 6.0 m. The fish travels
7.7 m horizontally before it hits the water
below.
What was the pelican’s initial speed? The
acceleration of gravity is 9.81 m/s
2. If the pelican was traveling at the same speed
but was only 1.6 m above the water, how
far would the fish travel horizontally before
hitting the water below?
To solve these problems, we can use the principles of projectile motion. In both scenarios, the horizontal motion of the fish is independent of its vertical motion.
Let's start with the first question: what was the pelican's initial speed?
In the first scenario, the fish is dropped from a height of 6.0 m, and it travels 7.7 m horizontally before hitting the water. We can find the time it takes for the fish to fall using the formula:
h = (1/2) * g * t^2
where h is the height, g is the acceleration due to gravity, and t is the time of flight.
Substituting the given values:
6.0 = (1/2) * 9.81 * t^2
Simplifying the equation:
t^2 = (2 * 6.0) / 9.81
t^2 = 1.22
t ≈ 1.11 s
Now that we have the time, we can calculate the initial vertical velocity using:
v = g * t
v = 9.81 * 1.11
v ≈ 10.89 m/s
Since the horizontal motion is independent, the initial horizontal velocity is equal to the fish's horizontal distance divided by the time:
v_h = d_h / t
v_h = 7.7 / 1.11
v_h ≈ 6.94 m/s
Therefore, the pelican's initial speed is the resultant of its horizontal and vertical velocities:
v_initial = √(v_h^2 + v^2)
v_initial = √(6.94^2 + 10.89^2)
v_initial ≈ 13.10 m/s
So the pelican's initial speed was approximately 13.10 m/s.
Now let's move on to the second question: how far would the fish travel horizontally if the pelican was only 1.6 m above the water?
In this scenario, the fish is dropped from a height of 1.6 m. We can use the same formula for time:
1.6 = (1/2) * 9.81 * t^2
Solving for t:
t^2 = (2 * 1.6) / 9.81
t^2 = 0.326
t ≈ 0.57 s
Now, using the same formula as before, we can find the horizontal distance:
d_h = v_h * t
d_h = 6.94 * 0.57
d_h ≈ 3.96 m
Therefore, if the pelican was only 1.6 m above the water, the fish would travel approximately 3.96 m horizontally before hitting the water.