Question

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?]

Answer 1

Answer:

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:

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