A helicopter 8.40 m above the ground and descending at 4.50 m/s drops a package from rest (relative to the helicopter). We have chosen the positive positive direction to be upward. The package falls freely.
a. Just as it hits the ground, find the y -component of the velocity of the package relative to the helicopter.
b. Just as it hits the ground, find the y -component of the velocity of the helicopter relative to the package.
To solve this problem, we can use the equations of motion for vertically falling objects. Let's break down each part of the question and solve it step by step.
a. To find the y-component of the velocity of the package relative to the helicopter just as it hits the ground, we can use the equation:
v = v₀ + at
Where:
v is the final velocity
v₀ is the initial velocity (which is 0, as the package is at rest relative to the helicopter)
a is the acceleration due to gravity (-9.8 m/s², assuming downwards is positive)
t is the time it takes for the package to hit the ground.
First, we need to find the time it takes for the package to hit the ground. We can use the equation:
y = y₀ + v₀t + (1/2)at²
Where:
y is the final position (which is 0, as the ground is the reference point)
y₀ is the initial position (8.40 m above the ground)
v₀ is the initial velocity (0)
a is the acceleration due to gravity (-9.8 m/s²)
t is the time it takes for the package to hit the ground.
Rearranging the equation, we have:
0 = 8.40 + 0t + (1/2)(-9.8)t²
Simplifying the equation, we get:
4.9t² = 8.40
Dividing both sides by 4.9, we get:
t² = 8.40 / 4.9
Taking the square root of both sides, we find:
t ≈ 1.43 s
Now that we have the time, we can find the final velocity using the first equation:
v = 0 + (-9.8)(1.43)
Calculating that, we get:
v ≈ -14.04 m/s
So, just as it hits the ground, the y-component of the velocity of the package relative to the helicopter is approximately -14.04 m/s (downward).
b. To find the y-component of the velocity of the helicopter relative to the package just as it hits the ground, we can use the fact that the package is falling freely with an acceleration due to gravity of -9.8 m/s².
Since both the helicopter and the package are subject to the same acceleration, their relative velocity in the y-direction will be constant.
Therefore, the y-component of the velocity of the helicopter relative to the package just as the package hits the ground will also be approximately -14.04 m/s (downward).
Note: The negative sign indicates a downward direction, according to the chosen positive direction being upward.