The drawing shows an airplane moving horizontally with a constant velocity of 115 m/s at an altitude of 1050 m. The direction to the right has been chosen as the +x direction, and upward is the +y direction. The plane releases a "care package" that falls to the ground along a curved trajectory. This situation was analyzed in Example 3. Suppose now that this plane is traveling with a horizontal velocity of +248 m/s. If all other factors remain the same, determine the time required for the package to hit the ground.
if all other factors are the same, then it will take the same amount of time to hit the ground, since the direction and height are the same. The horizontal speed makes no difference in how fast it falls.
To determine the time required for the package to hit the ground, we need to analyze the vertical motion of the package separately.
Given:
- Initial altitude (y₀) = 1050 m
- Acceleration due to gravity (g) = 9.8 m/s² (downward)
We can use the equation of motion for vertical motion:
y = y₀ + v₀y*t + (1/2)*a*t²
Since the initial velocity in the vertical direction (v₀y) is 0 (the package is released from rest), the equation simplifies to:
y = y₀ + (1/2)*a*t²
Substituting the values:
y = 1050 m
a = -9.8 m/s² (negative because it is downward)
t = time (unknown)
We are looking for the time when the package hits the ground, which means y = 0.
0 = 1050 + (1/2)*(-9.8)*t²
Simplifying the equation:
0 = 1050 - 4.9*t²
Rearranging the equation:
4.9*t² = 1050
Dividing both sides by 4.9:
t² = 1050 / 4.9
t² = 214.2857
Taking the square root of both sides:
t ≈ √214.2857
t ≈ 14.63 seconds
Therefore, the time required for the package to hit the ground is approximately 14.63 seconds.
To determine the time required for the package to hit the ground, we can use the kinematic equations of motion.
First, let's consider the vertical motion of the package. The initial vertical position is 1050 m, and the final vertical position is 0 m (since it hits the ground). The acceleration due to gravity (g) acts downwards. The initial vertical velocity is 0 m/s, as the package is initially just released. We need to find the time it takes for the package to reach the ground.
Using the equation for vertical displacement:
y = y0 + v0y * t + (1/2) * a * t^2
Where:
y = final vertical position (0 m)
y0 = initial vertical position (1050 m)
v0y = initial vertical velocity (0 m/s)
a = acceleration due to gravity (-9.8 m/s^2, negative because it acts downwards)
t = time
Plugging in the known values, we get:
0 = 1050 + 0 * t + (1/2) * (-9.8) * t^2
Simplifying the equation, we have:
4.9t^2 = 1050
Now, let's solve for t by isolating the variable:
t^2 = 1050 / 4.9
t^2 = 214.29
t ≈ √214.29
t ≈ 14.63 s
Therefore, it takes approximately 14.63 seconds for the package to hit the ground.