Physics, 28.09.2020 14:01 brandydailey24pe8r24

A helicopter is ascending vertically with a speed of 5.32 m/s . At a height of 120 m 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 v0 for the package?]

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Answer from: kamkamkab

Approximately $5.52\; \rm s$ (assuming that $g = -9.81\; \rm m\cdot s^{-2}$ .)

Explanation:

From the perspective of an observer on the ground, the package was initially moving with the helicopter. The initial velocity of that package ( v_0 ) would be the same as the velocity of the helicopter. In other words, $v_0 = 5.32\; \rm m \cdot s^{-1}$ .

Other quantities:

$a = g \approx -9.81\; \rm m \cdot s^{-2}$ . (The

here is negative because the acceleration due to gravity points downwards.) $x = -120\; \rm m \cdot s^{-1}$ (The position of the package would have changed by $120\; \rm m$ by the time it reached the ground. The negative sign is because this position change also points downwards.)

There are two ways to find the time required for the package to reach the ground after it was dropped.

The first approach makes use of the SUVAT equation:

$\displaystyle x = \frac{1}{2}\, a\, t^2 - v_0\, t$ .

For this question, , , and v_0 are already known. However, this approach would require solving a quadratic equation where the coefficient for the first-order term is non-zero.

An alternative approach is to solve for v_1 , the velocity of the package right before it reaches the ground. From the SUVAT equation that does involve time:

$\displaystyle {v_1}^{2} - {v_0}^{2} = 2\, a\, x$ .

Again, , , and v_0 are already known. Solve for v_1 :

$\begin{aligned}v_1 &= \sqrt{2\, a\, x + {v_0}^2} \\ &\approx \sqrt{2\times (-9.81) \times (-120) + 5.32^2} \approx -48.81293\; \rm m \cdot s^{-1}\end{aligned}$ .

Calculate time from the change in velocity:

$\displaystyle t = \frac{v_1 - v_0}{a} \approx 5.52\; \rm s$ .

Answer from: Quest

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Answer from: Quest

air exerts forces on falling objects near earth’s surface.

explanation:

the term 'free falls' refer to object that are falling towards the earth, in a situation in which gravity is the only force acting on the object.

for an object in free fall, the acceleration is constant (g, the gravitational acceleration), and so the velocity of the object constantly increases.

for objects near earth's surface, there is always another force acting on the object: the resistance due to the air. this force is opposed to the direction of fall of the object, and it is proportional to the velocity of the object, so as the object falls down, this force increases up to a point when it balances the force of gravity, and the object continues its motion at constant velocity (called terminal velocity). since in this case gravity is not the only force acting on the object, we are not in a situation of free fall.

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A helicopter is ascending vertically with a speed of 5.32 m/s . At a height of 120 m above the Earth...