A pelican flying along a horizontal path drops a fish from a height of 5.5 m. The fish travels 7.0 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.
7.0 m/(time to fall 5.5 m)
Time to fall Y = = sqrt (2Y/g)
= 1.059 s for Y = 5.5 m
Now use the first equation
To find the pelican's initial speed, we can use the kinematic equation for vertical motion:
h = (1/2) * g * t^2
where h is the vertical distance, g is the acceleration due to gravity, and t is the time it takes for the fish to fall to the water.
First, let's find the time it takes for the fish to fall. Since the pelican dropped the fish from a height of 5.5 m and it falls vertically, we can use the formula:
h = (1/2) * g * t^2
Rearranging the equation to solve for t:
t^2 = (2 * h) / g
t^2 = (2 * 5.5) / 9.81
t^2 = 1.12
t = √1.12
t ≈ 1.06 seconds (rounded to two decimal places)
Next, we can use the horizontal distance traveled by the fish to find the initial speed of the pelican. The horizontal motion of the fish is independent of its vertical motion, so we can use the equation:
d = v * t
where d is the horizontal distance traveled, v is the initial velocity, and t is the time.
Rearranging the equation to solve for v:
v = d / t
v = 7.0 / 1.06
v ≈ 6.6 m/s (rounded to one decimal place)
Therefore, the pelican's initial speed was approximately 6.6 m/s.