A helicopter is ascending vertically with a speed of 5.40m/s at a height of 105m above the earth, a package is dropped from the helicopter. How much time does it take for the package to reach the ground? (Hint: What is Vo for the package?)
The Correct Answer and Explanation is :
To solve this problem, we need to calculate the time it takes for the package to hit the ground after being dropped from a helicopter. Let’s break the problem down step by step.
Given Data:
- Initial velocity of the package (Vo): The package is dropped from a helicopter moving upward at 5.40 m/s. Therefore, the initial velocity of the package (relative to the ground) is the same as the velocity of the helicopter: ( V_o = 5.40 \, \text{m/s} ) (upward).
- Height (h): The height of the package when released is ( h = 105 \, \text{m} ).
- Acceleration due to gravity (g): The acceleration due to gravity is ( g = 9.81 \, \text{m/s}^2 ) (downward).
- Final position: We want to determine the time it takes for the package to reach the ground, so the final position will be 0 meters.
Relevant Equation:
We can use the kinematic equation for vertical motion, which relates position, initial velocity, acceleration, and time:
[
y = V_o \cdot t + \frac{1}{2} a \cdot t^2
]
Where:
- ( y ) is the vertical displacement (in this case, -105 meters because the package falls 105 meters to the ground),
- ( V_o ) is the initial velocity (5.40 m/s, upwards),
- ( a ) is the acceleration due to gravity (-9.81 m/s², downward),
- ( t ) is the time we need to solve for.
Step 1: Set up the equation
Substitute the known values into the equation:
[
-105 = (5.40) \cdot t + \frac{1}{2} (-9.81) \cdot t^2
]
[
-105 = 5.40 \cdot t – 4.905 \cdot t^2
]
Step 2: Solve the quadratic equation
This is a quadratic equation of the form:
[
0 = -4.905 \cdot t^2 + 5.40 \cdot t + 105
]
We can solve it using the quadratic formula:
[
t = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}
]
Where:
- ( a = -4.905 ),
- ( b = 5.40 ),
- ( c = 105 ).
Substituting into the quadratic formula:
[
t = \frac{-5.40 \pm \sqrt{(5.40)^2 – 4(-4.905)(105)}}{2(-4.905)}
]
First, calculate the discriminant:
[
\text{Discriminant} = (5.40)^2 – 4(-4.905)(105) = 29.16 + 2052.06 = 2081.22
]
Now solve for ( t ):
[
t = \frac{-5.40 \pm \sqrt{2081.22}}{-9.81}
]
[
t = \frac{-5.40 \pm 45.64}{-9.81}
]
Step 3: Calculate the two possible solutions
The two possible solutions are:
[
t_1 = \frac{-5.40 + 45.64}{-9.81} = \frac{40.24}{-9.81} \approx 4.10 \, \text{seconds}
]
[
t_2 = \frac{-5.40 – 45.64}{-9.81} = \frac{-51.04}{-9.81} \approx 5.21 \, \text{seconds}
]
Since the time of flight must be a positive value, we take the solution ( t_1 = 4.10 \, \text{seconds} ).
Final Answer:
It takes approximately 4.10 seconds for the package to reach the ground.
Explanation:
- The package starts with an upward velocity of 5.40 m/s and is subjected to downward acceleration due to gravity at 9.81 m/s².
- As the package moves upward initially, it slows down due to gravity until it reaches a maximum height.
- After that, it starts accelerating downward and eventually hits the ground.
- The kinematic equation is used to model this motion and calculate the time it takes for the package to fall to the ground, considering both the initial upward velocity and the downward gravitational force.