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. Answer:

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

    Explanation:

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

    Other quantities:

    • [tex]a = g \approx -9.81\; \rm m \cdot s^{-2}[/tex]. (The [tex]g[/tex] here is negative because the acceleration due to gravity points downwards.)
    • [tex]x = -120\; \rm m \cdot s^{-1}[/tex] (The position of the package would have changed by [tex]120\; \rm m[/tex] 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 [tex]t[/tex] required for the package to reach the ground after it was dropped.

    The first approach makes use of the SUVAT equation:

    [tex]\displaystyle x = \frac{1}{2}\, a\, t^2 – v_0\, t[/tex].

    For this question, [tex]x[/tex], [tex]a[/tex], and [tex]v_0[/tex] 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 [tex]v_1[/tex], the velocity of the package right before it reaches the ground. From the SUVAT equation that does involve time:

    [tex]\displaystyle {v_1}^{2} – {v_0}^{2} = 2\, a\, x[/tex].

    Again, [tex]x[/tex], [tex]a[/tex], and [tex]v_0[/tex] are already known. Solve for [tex]v_1[/tex]:

    [tex]\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}[/tex].

    Calculate time from the change in velocity:

    [tex]\displaystyle t = \frac{v_1 – v_0}{a} \approx 5.52\; \rm s[/tex].

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