an eagle flying in a horizontal path drops a twig from a height of 8.2 m. The twig travels 12 m horizontally before it hits the ground. What was the eagle's initial speed?
7.1 m/s
To find the initial speed of the eagle, we can use the horizontal distance traveled by the twig before hitting the ground and the height from which the twig was dropped.
Let's break the problem down into two components: horizontal and vertical.
1. Horizontal Component:
The horizontal component of the eagle's motion remains the same throughout, as there are no horizontal forces acting on it. Therefore, the initial horizontal velocity (Vx) of the eagle remains constant.
Vx = ?
2. Vertical Component:
The vertical motion of the twig can be analyzed using the equations of motion under constant acceleration due to gravity (g = 9.8 m/s^2). We consider the vertical distance traveled by the twig (8.2 m) and the time it takes to reach the ground.
Given:
Initial vertical position (y0) = 8.2 m
Final vertical position (y) = 0 m (height when it hits the ground)
Acceleration due to gravity (g) = 9.8 m/s^2
Time taken to reach the ground (t) = ?
Using the kinematic equation:
y = y0 + Vyt - (1/2)gt^2
Substituting the known values:
0 = 8.2 + Vy(t) - (1/2)(9.8)t^2
We know that Vy (vertical component of velocity) is 0 at the highest point of the trajectory, as the twig only has a horizontal motion from there onwards.
Simplifying the equation:
-8.2 = -(1/2)(9.8)t^2
Solving for time:
t^2 = (2 * 8.2) / 9.8
t^2 = 16.4 / 9.8
t^2 = 1.673
t = √1.673
t ≈ 1.293 seconds
Now that we have the time taken by the twig to reach the ground, we can calculate the horizontal velocity (Vx) using the horizontal distance traveled by the twig before hitting the ground (12 m).
Vx = distance / time
Vx = 12 m / 1.293 s
Vx ≈ 9.28 m/s
Therefore, the initial horizontal speed (Vx) of the eagle is approximately 9.28 m/s.