A pelican flying along a horizontal path drops

a fish from a height of 3.3 m. The fish travels
7.6 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

d = 7.6 m.

h = 0.5g*t^2 = 3.3 m. Solve for t.

Xo = d/t m/s = Initial Hor. velocity

To find the initial speed of the pelican, you can use the equations of motion. First, let's identify the known values from the problem:

Height (vertical distance) = 3.3 m
Horizontal distance = 7.6 m
Acceleration due to gravity = 9.81 m/s^2

We can use the equation of motion in the vertical direction to find the time it takes for the fish to fall:

h = (1/2)gt^2

where:
h = height (3.3 m)
g = acceleration due to gravity (9.81 m/s^2)
t = time

Rearranging the equation, we can solve for t:

t = sqrt(2h/g)

Substituting the given values, we get:

t = sqrt(2 * 3.3 / 9.81)
t ≈ 0.64756 s (rounded to 5 decimal places)

Now that we have the time it takes for the fish to fall, we can use the equation of motion in the horizontal direction to find the initial velocity:

s = ut

where:
s = horizontal distance (7.6 m)
u = initial velocity (what we want to find)
t = time (0.64756 s)

Rearranging the equation, we can solve for u:

u = s / t

Substituting the given values, we get:

u = 7.6 / 0.64756
u ≈ 11.736 m/s (rounded to 3 decimal places)

Therefore, the pelican's initial speed was approximately 11.736 m/s.

To find the initial speed of the pelican, we can use the kinematic equation:

s = ut + (1/2)at^2

where:
s = displacement (horizontal distance traveled by the fish) = 7.6 m
u = initial velocity (pelican's initial speed)
t = time taken (unknown)
a = acceleration (acceleration due to gravity) = -9.81 m/s^2 (negative because it acts downwards)

We can rearrange the equation to solve for the initial velocity:

u = (s - (1/2)at^2)/t

First, let's find the time taken by the fish to travel 7.6 m horizontally. Since there is no vertical acceleration (we only have the horizontal displacement), the time taken will be the same as if the fish were in free fall vertically.

We can use the formula:

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

where:
s = vertical displacement (height from which the fish is dropped) = 3.3 m
g = acceleration due to gravity = 9.81 m/s^2 (positive because it acts downwards)

Plugging in the values:

3.3 = 0t + (1/2)(9.81)t^2

Simplifying:

4.905t^2 = 3.3

t^2 = 3.3 / 4.905

t^2 = 0.672515

t ≈ √0.672515

t ≈ 0.819 s (rounded to three decimal places)

Now, we can substitute the value of t into the equation for initial velocity:

u = (s - (1/2)at^2)/t

u = (7.6 - (1/2)(-9.81)(0.819^2))/0.819

u = (7.6 + 0.8137)/0.819

u = 8.4137/0.819

u ≈ 10.26 m/s (rounded to two decimal places)

Therefore, the pelican's initial speed was approximately 10.26 m/s.