. A package is dropped from a helicopter moving upward at 5 m/s. If it takes the package 15 s to strike the
ground, how high above the ground was the package when it was released? Assume that air resistance is
negligible
h + 5t - 4.9t^2 = 0
now plug in t=15 and solve for h
We can solve this problem using the equations of motion. The motion of the package can be divided into two parts: the upward motion of the helicopter and the downward motion of the package.
Let's consider the upward motion of the helicopter first. We know that it is moving upward at a constant velocity of 5 m/s. The time taken for the package to strike the ground is 15 s. Since the package is released from the helicopter, its initial velocity is the same as that of the helicopter, which is 5 m/s.
Using the equation of motion, we can calculate the distance covered during the upward motion of the helicopter:
distance = initial velocity × time + (1/2) × acceleration × time^2
Since the helicopter's velocity is constant and directed upward, the acceleration is zero. Therefore, the equation simplifies to:
distance = initial velocity × time
distance = 5 m/s × 15 s = 75 m
So, during the upward motion of the helicopter, it covers a distance of 75 meters.
Now, let's consider the downward motion of the package. The package is being dropped, so its initial velocity is zero. The acceleration due to gravity, denoted by "g," is approximately 9.8 m/s^2 (assuming standard gravity).
Using the equation of motion, we can calculate the distance covered during the downward motion of the package:
distance = (1/2) × acceleration × time^2
distance = (1/2) × 9.8 m/s^2 × (15 s)^2
distance ≈ 1102.5 m
Now, we can find the height above the ground when the package was released by subtracting the distance covered during the upward motion of the helicopter from the distance covered during the downward motion of the package:
height = distance downward - distance upward
height ≈ 1102.5 m - 75 m
height ≈ 1027.5 m
Therefore, the package was approximately 1027.5 meters above the ground when it was released.
To find the height above the ground at which the package was released, we can use the equations of motion.
We need to determine two key pieces of information: the initial velocity and the time it took for the package to strike the ground.
Given:
Initial upward velocity of the helicopter, v = 5 m/s
Time taken to strike the ground, t = 15 s
To find the height, we can use the equation:
h = ut + (1/2)gt^2
Where:
h is the height
u is the initial velocity
g is the acceleration due to gravity (approximately 9.8 m/s^2)
In this case, the package is moving upwards, so the initial velocity is positive (+5 m/s). The acceleration due to gravity acts in the downward direction, so it is negative (-9.8 m/s^2).
Substituting the given values into the equation, we have:
h = (5 m/s) * (15 s) + (1/2) * (-9.8 m/s^2) * (15 s)^2
Simplifying,
h = 75 m - (1/2) * 9.8 * 225
h = 75 m - (1/2) * 9.8 * 225
h = 75 m - 441.75 m
h = -366.75 m
Since height cannot be negative, we take the magnitude of the value:
h ≈ 366.75 m
Therefore, the package was approximately 366.75 meters above the ground when it was released.