THE CHOPPER RISES FROM REST ON THE GROUND VERTICALLYUPWARDS WITH A CONSTANT ACCLERATION g/8. A FOOD PACKET IS DROPPED WHEN IT HAS RISEN TO A HEIGHT h. SHOW THAT THE TIME TAKEN BY THE PACKET TO REACH THE GROUND IS 2 * ROOT UNDER h/g

Let us take the ground frame as the reference frame.

Let u be the initial velocity of the chopper and v be its velocity at height h.

According to the equations of the laws of motion,

v²-(0)²=2(g/8)h

v²=gh/4

v=√(gh)/2

So,the initial velocity of the food pachet when it is dropped from the height 'h' is √(gh)/2 in upward direction(one should take care of the sign conventions in these types of problems)

When the object is dropped,it has an accn. g in downward direction.

Let the time taken by the packet to reach the ground be 't'.

According to the equations of the laws of motion,

-h={√(gh)/2}t+(1/2)(-g)t²

(g/2)t² - {√(gh)/2}t-h=0

Solving the quadratic equation,we get

t=2(√h/g),-√(h/g)

Since time cannot be negative,

time taken by the packet to reach the ground (t)=2√(h/g)