A pelican flying along a horizontal path drops a fish from a height of 4.2 m. The fish travels 9.1 m horizontally before it hits the water below.
What was the pelican’s initial speed? The
acceleration of gravity is 9.81 m/s2.
To find the pelican's initial speed, we can use the kinematic equations of motion. In this case, we'll use the horizontal motion equation, which relates the initial speed (Vi), final speed (Vf), acceleration (a), and displacement (d):
d = (Vi + Vf) / 2 * t
In this scenario, the displacement is the horizontal distance the fish travels before hitting the water, which is 9.1 m. We'll consider the acceleration to be zero in the horizontal direction because there are no external forces acting on the fish in that direction.
Since we know that the fish travels 9.1 m horizontally, and there is no acceleration in that direction, we can determine the time it takes for the fish to hit the water using the formula:
d = Vi * t
Solving for t:
t = d / Vi
Substituting the values:
t = 9.1 m / Vi
Now we can substitute the value of t back into the horizontal motion equation to find Vi:
9.1 m = (Vi + 0) / 2 * (9.1 m / Vi)
Multiplying both sides by 2 * Vi:
18.2 m * Vi = 9.1 m
Simplifying:
Vi = 9.1 m / 18.2 m
Vi = 0.5 m/s
Therefore, the pelican's initial speed was 0.5 m/s.
To determine the initial speed of the pelican, we can use the equations of motion for horizontal (x) and vertical (y) directions.
In the vertical direction, we have:
y = y0 + v0y * t + 0.5 * a * t^2
Where:
y = vertical displacement (4.2 m)
y0 = initial vertical position (0 m)
v0y = initial vertical velocity (unknown)
a = acceleration due to gravity (-9.81 m/s^2)
t = time taken to reach the water (unknown)
Since the pelican drops the fish, the final vertical position (y) is at the water level (y = 0). Thus, the equation simplifies to:
0 = 0 + v0y * t + 0.5 * (-9.81) * t^2
Simplifying further, we have:
4.9 * t^2 - v0y * t = 0
Dividing the equation by t, we get:
4.9 * t - v0y = 0
v0y = 4.9 * t
Now, let's consider the horizontal direction. The fish travels 9.1 m horizontally before hitting the water. The time taken (t) is the same for both vertical and horizontal directions since the pelican dropped the fish without exerting any horizontal force.
Using the equation of motion in the horizontal direction (assuming a constant horizontal velocity):
x = x0 + v0x * t
Where:
x = horizontal displacement (9.1 m)
x0 = initial horizontal position (0 m)
v0x = initial horizontal velocity (unknown)
t = time taken (same for both horizontal and vertical directions)
Simplifying the equation, we have:
9.1 = 0 + v0x * t
v0x = 9.1 / t
Since v0x is the initial speed of the pelican, we can substitute it as v0 in the equation:
v0 = sqrt(v0x^2 + v0y^2)
Substituting v0x and v0y:
v0 = sqrt((9.1 / t)^2 + (4.9 * t)^2)
Now, to find the initial speed, we need to solve for t so that the fish will travel 9.1 m horizontally.
Using the equation:
x = x0 + v0x * t
9.1 = 0 + v0x * t
t = 9.1 / v0x
Substituting this value of t back into the equation for v0:
v0 = sqrt((9.1 / (9.1 / v0x))^2 + (4.9 * (9.1 / v0x))^2)
Simplifying further:
v0 = sqrt((v0x)^2 + (4.9 * 9.1)^2 / (v0x)^2)
Now, we can solve for v0 using this equation.