A pelican flying along a horizontal path drops a fish from a height of 5.1 m. The fish travels 6.1 m horizontally before it hits the water below. What was the pelican’s initial speed? The acceleration of gravity is 9.81 m/s2 . Answer in units of m/s
To find the pelican's initial speed, we can use the principles of projectile motion. The key is to recognize that the vertical motion (dropping the fish) and horizontal motion (the fish traveling horizontally) are independent of each other.
We'll begin by finding the time it takes for the fish to fall 5.1 m vertically. We can use the equation of motion for vertical displacement:
y = vt + (1/2)gt^2
where y is the vertical displacement (5.1 m), v is the initial vertical velocity (which is zero since the fish is dropped), g is the acceleration due to gravity (-9.81 m/s^2), and t is the time.
Since the initial vertical velocity is zero, the equation simplifies to:
y = (1/2)gt^2
Rearranging the equation to solve for t:
t = sqrt(2y/g)
Substituting the given values:
t = sqrt(2 * 5.1 / 9.81)
t ≈ 0.32 s
Now, we can determine the horizontal distance travelled by the fish using the horizontal motion equation:
x = vt
where x is the horizontal distance (6.1 m), v is the initial horizontal velocity, and t is the time we just calculated.
Rearranging the equation to solve for v:
v = x / t
Substituting the known values:
v = 6.1 / 0.32
v ≈ 19.06 m/s
Therefore, the pelican's initial speed is approximately 19.06 m/s.