. A package is dropped from a helicopter moving upward at 1.5 m/s. If it takes 1

Question

. A package is dropped from a helicopter moving upward at 1.5 m/s. If it takes 16.0 s before the package strikes the ground, how high above the ground was the package when it was released if air resistance is negligible? . A package is dropped from a helicopter moving upward at 1.5 m/s. If it takes 16.0 s before the package strikes the ground, how high above the ground was the package when it was released if air resistance is negligible? . A package is dropped from a helicopter moving upward at 1.5 m/s. If it takes 16.0 s before the package strikes the ground, how high above the ground was the package when it was released if air resistance is negligible?

Explanation / Answer

Let h be the height of the helicopter when the package was dropped. Also, let's assume downward direction to be positive (just a convention).

Given, u = initial speed of package = upwards speed of helicopter = -1.5 m/s (upwards hence negative), t = 16.0 s, a = g = 9.81 m/s2, h = ?

From the second equation of uniform motion -

s = ut + (1/2)at2

Therefore,

h = -1.5 t + ((1/2) x 9.81 x t2) = (-1.5 x 16) + (0.5 x 9.81 x 162) = 1231.68 m = 1.23 km

Thus, the helicopter was nearly 1.23 km above the ground when the package was dropped.

Related Questions

Navigate

• Browse (All)
• Browse #
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.