Aflying 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

given,

vertical speed of the dragon = 6 m/s

height of dragon above ground = 30 m

a) maximum height of the ball.

 At the topmost point velocity is equal to zero

 Using equation of motion

           h = 1.84 m

maximum height of the egg above ground

  H = 30 + 1.84 = 31.84 m

b) time taken by the ball to reach the ground

 4.9 t² - 6 t - 30 = 0

on solving the above equation

  t = 3.16 s (neglecting the negative sign)

c) egg velocity just before it hit the ground

   using equation of motion

      v² = u² + 2 g h

      v² = 0 + 2 x 9.8 x 31.84

      v = √624.068

      v = 24.98 m/s

d) The graphs are attached below.

   In v-t graph velocity at the top most point is zero and after that it is negative

  In a-t graph only acceleration due to gravity is acting in the negative direction.

the answer is 3.1 his answer is wrong mines is right

higher potential

explanation:

because it is in a high speed so it must move into a of high potential