A helicopter 7.60m above the ground and descending at 3.90m/s drops a package from rest (relative to the helicopter). We have chosen the positive positive direction to be upward. The package falls freely.
part A- Just as it hits the ground, find the velocity of the package relative to the helicopter.
Vph=?m/s
part B- Just as it hits the ground, find the velocity of the helicopter relative to the package.
Vph=?m/s
ok for Part A i did 3.9^2+2*9.8*7.6= 164.17 the sqaure root it = 12.812 and i put -12.8 and i got it wrong the formula i used is Vf^2= V1^2+ 2gh i need help please
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I think your answer is correct.
yea i put in masterphyics hw and it said its wrong i did it 3 times this question is driving me crazy
are you paying for "masterphysics" service? Hmmm. Complain.
yes i did pay but they are not going to help
anyway can you help me solve this problem please
To solve this problem, let's break it down into two parts: the motion of the package relative to the ground and the motion of the helicopter relative to the ground.
Part A: Velocity of the package relative to the helicopter
Since the package is dropped from rest relative to the helicopter, its initial velocity with respect to the helicopter is 0 m/s. We want to find the velocity of the package when it hits the ground, relative to the helicopter.
To determine this, we can use the equation for the final velocity, Vf, of an object in free fall:
Vf^2 = Vi^2 + 2ad
Where:
Vf = final velocity (which is what we want to find)
Vi = initial velocity (0 m/s in this case)
a = acceleration due to gravity (approximately 9.8 m/s^2)
d = distance (in this case, the height from which the package was dropped, 7.60 m)
Plugging in the known values:
Vf^2 = 0^2 + 2*9.8*7.6
Vf^2 = 0 + 147.84
Vf^2 = 147.84
To find Vf, we take the square root of both sides:
Vf = √147.84
Vf ≈ 12.15 m/s
Therefore, the velocity of the package relative to the helicopter just as it hits the ground is approximately 12.15 m/s upward (taking upward as the positive direction).
Part B: Velocity of the helicopter relative to the package
We want to find the velocity of the helicopter relative to the package just as the package hits the ground.
Since the package falls freely, its final velocity relative to the ground is the same as the velocity of the helicopter relative to the ground.
Therefore, the velocity of the helicopter relative to the package just as the package hits the ground is also approximately 12.15 m/s upward (taking upward as the positive direction).
Note: It is important to be consistent with the sign convention. In this case, we are considering upward as positive and downward as negative.