Question

A flying dragon is rising vertically at a constant speed of 6.0m/s. When the dragon is 30.0m above the ground, the rider on its back drops a small golden egg which, subsequently, is in free fall.a) What is the maximum height above the ground reached by the egg?b) How long after its release does the egg hit the ground?c) What is the egg’s velocity immediately before it hits the ground?d) Sketch, qualitatively, position, velocity, and acceleration of the egg as functions of time.

Answer 1

Answer:

a)  y = 31.84 m , b)  t₂ = 3.16 s , c) v = -24.97 m / s

Explanation:

a) We can solve this exercise using kinematic expressions

At the highest point the speed is zero

           v² = v₀² - 2 g (y - y₀)

           y = y₀ + v₀² / 2 g

           y = 30 +6²/2 9.8

           y = 31.84 m

b) The time for the egg to reach the ground (y = 0 m)

            y = y₀ + v₀ t - ½ g t²

            0 = 30 + 6 t - ½ 9.8 t²

            4.9 t² - 6t - 30 = 0

            t² - 1.22 t - 6.12 = 0

Let's solve the second degree equation

            t = [1.22 ±√ (1.22² - 4 (-6.12))] / 2

            t = (1.22 ± 5.1) / 2

            t₁ = -1.94 s

            t₂ = 3.16 s

We fear positive time

c) The speed at this point

              v = v₀ - gt

              v = 6 - 9.8 3.16

              v = -24.97 m / s

The negative sign indicates that the speed is down