A pelican flying along a horizontal path drops

a fish from a height of 3.1 m. The fish travels
6.6 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
.
Answer in units of m/s

To find the pelican's initial speed, we can use the equations of motion.

We know that the fish traveled a horizontal distance of 6.6 m and dropped from a height of 3.1 m. Let's assume the initial speed of the pelican is "v" m/s.

Using the first equation of motion, we can find the time it takes for the fish to hit the water:

h = (1/2) * g * t^2

where h is the height, g is the acceleration due to gravity, and t is the time.

Plugging in the values, we have:

3.1 = (1/2) * 9.81 * t^2

Simplifying the equation, we have:

3.1 = 4.905 * t^2

Dividing both sides by 4.905, we get:

t^2 = 0.632384

Taking the square root of both sides, we find:

t = 0.795506 s

Now, we can use the second equation of motion to find the initial horizontal velocity of the fish:

s = v * t

where s is the horizontal distance and v is the initial velocity.

Plugging in the values, we have:

6.6 = v * 0.795506

Solving for v, we get:

v = 6.6 / 0.795506

v ≈ 8.28596 m/s

Therefore, the pelican's initial speed is approximately 8.28596 m/s.

To find the pelican's initial speed, we can use the principles of projectile motion.

First, let's consider the vertical motion of the fish. We can use the equation of motion for vertical motion:

h = ut + (1/2)gt^2

Where:
h = vertical distance traveled (3.1 m)
g = acceleration due to gravity (-9.81 m/s^2)
t = time taken to reach the water

Since the fish is dropped vertically, its initial velocity in the vertical direction (u) is 0. We can rearrange the equation and solve for t:

3.1 = 0*t + (1/2)(-9.81)*t^2
3.1 = (-4.905)*t^2

Simplifying the equation, we get:
t^2 = 3.1 / (-4.905)
t^2 = -0.632
Note that we ignore the negative value since time cannot be negative.

Taking the square root of both sides:
t = √(-0.632)
t = √(0.632)i
where i represents the imaginary unit.

Since we are considering real-world motion, it means there is no real value of time that satisfies this equation. Therefore, the initial vertical velocity is not sufficient to solve the problem.

Now, let's consider the horizontal motion of the fish. We can use the equation of motion for horizontal motion:

s = ut

where:
s = horizontal distance traveled (6.6 m)
u = initial horizontal velocity (same as pelican's initial speed)
t = time taken to reach the water

We can rearrange the equation and solve for u:

6.6 = u*t
Since we couldn't find the real value of t, we cannot solve for u.

Therefore, given the information provided, we cannot determine the pelican's initial speed.