a helicopter is ascending vertically with a speed of 5.20 m/s with package of negligible size attached to the bottom of the helicopter, when the helicopter is 125 m above the ground the package is released and the helicopter continues to ascend at its same speed. air resistance is negligible in this problem.(a)what is the displacement of the helicopter 5 seconds after the package has been released?(b)what is the displacement of the package 5.0 seconds after it has been released?(c)how far apart are the bottom of the helicopter and the package after 5.0 seconds?
a. displacement= 125+5.20*5 above ground
b. s=125+5.2*5+ 1/2 g t^2 (t=5, g is =9.8m/s^2)
c. subtract.
(a) 125 m + (5.20 m/s * 5 s)
(b) [-4.9 m/s^2 * (5 s)^2] + (5.20 m/s * 5 s) + 125 m
(c) result (a) - result (b)
To solve this problem, we will use the kinematic equations of motion.
(a) To find the displacement of the helicopter 5 seconds after the package has been released, we need to calculate the vertical distance traveled by the helicopter during this time.
The initial velocity of the helicopter is 5.20 m/s, and we want to find the displacement after 5 seconds. The acceleration due to gravity, g, is approximately 9.8 m/s².
We can use the equation:
displacement = initial velocity * time + 0.5 * acceleration * time²
Substituting in the values, we get:
displacement = 5.20 m/s * 5 s + 0.5 * 9.8 m/s² * (5 s)²
Simplifying the equation, we have:
displacement = 5.20 m/s * 5 s + 0.5 * 9.8 m/s² * 25 s²
Calculating, the displacement of the helicopter 5 seconds after the package has been released is approximately:
displacement = 26 m
(b) To find the displacement of the package 5 seconds after it has been released, we need to calculate the vertical distance traveled by the package during this time.
Since the package is released and has no initial velocity, the equation to use is:
displacement = 0.5 * acceleration * time²
Substituting in the values, we get:
displacement = 0.5 * 9.8 m/s² * (5 s)²
Simplifying the equation, we have:
displacement = 0.5 * 9.8 m/s² * 25 s²
Calculating, the displacement of the package 5 seconds after it has been released is approximately:
displacement = 122.5 m
(c) To find the distance between the bottom of the helicopter and the package after 5 seconds, we need to calculate the difference in their displacements.
The displacement of the package is already calculated as 122.5 m, and the displacement of the helicopter is calculated as 26 m. Taking the difference between them, we get:
distance = displacement of the package - displacement of the helicopter
Substituting in the values, we have:
distance = 122.5 m - 26 m
Calculating, the distance between the bottom of the helicopter and the package after 5 seconds is approximately:
distance = 96.5 m