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 helicopt

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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1. thucuc

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 $$g$$ 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 $$t$$ 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, $$x$$, $$a$$, 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, $$x$$, $$a$$, 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:

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