A gas balloon is powered by engines which can give it a constant but adjustable upward acceleration. At an instant of time when it is 509.5 m above the groundand has an upward velocity of 98.1 m/s the engines jam for 13 seconds.The engines restart after the period of 13 s . Now, to what constant acceleration should the engines be adjusted so as to reach the ground with zero velocity. Take the acceleration due to gravity as 9.81 m /s2 and assume that the gas balloon moves only in the vertical direction.

a)  0.453 m / s2 .
b) 10.263 m / s2 .
c)  9.810 m / s2 .
d) 11.810 m / s2  .


Let the points of interest be marked as A, B, C and D as shown in the figure. Let si, vi, ai and ti represent the displacement from the ground, instantaneous velocity, acceleration and time at the position i, where i = A, B, C and D.

Let sij and tij represent the displacement and the time taken between the points i and j, where i, j = A, B, C and D and i not equal to j.

At Point A ( Where the engines fail ):

Engines fail and the acceleration of the gas balloon is only due to the earths gravity.

    sA = 509.5 m.         vA = 98.1 m/s .
    aA = - 9.81 m/s2 .    tA = 0 s.

At Point B ( At te highest point of the path ):

Assuming that the acceleration till it reaches the highest point, B, is due to gravity alone; the time taken can be calculated from,

   vB = VA + a tB.
   0 = 98.1 + ( - 9.81) tB.
Hence,
   tB = 98.1 /9.81 = 10 s

Since 10 s is less than the time of jam, 13 s, our assumption above is justified.

From, vB2 - VA2 = 2 a sAB
   0 - ( 98.1)2 = 2 (-9.81)sAB
   sAB = ( 98.1)2 / 2( 9.81) = 490.5 m

Hence the maximum height above
the ground that it reaches is

   sA + sB = 509.5 + 490.5 = 1000 m.

At Point C ( Where the engines restart ) :

The point C is the point at which the engines restart. This point is 3 seconds ( 13 - 10 ) seconds away from point B. The displacement sBC and the velocity at C , vC, are calculated as follows

   sBC = vB tBC + 0.5 a tBC2.
   sBC = 0 + 0.5( - 9.81) (3)2 .
Hence,
   sBC = - 44.145 m.
And sC = 1000 - 44.145 = 955.855 m
   sCD = - 955.855 m

The velocity at C is given by
   vC = vB + a tBC = 0 + ( - 9.81) (3)
   vC = - 29.43 m/s.

At point D :

Since it just touches the ground at D ,  the velocity vD = 0 m/s.The required net acceleration, anet, can be calculated from the following equations.

   vD2 - vC2 = 2 (anet)(sC)
   0 - (29.43)2 = 2(anet)( - 955.855).
Hence,
   anet = (29.43)2 / 2(955.855) = 0.453 m /s2

Acceleration due to gravity is always present and is acting downwards at any instant of time.
Hence, the acceleration of the engines, ae, is given by
   anet = ae + g .
   ae = anet - g.
   ae = 0.453 - ( - 9.81) = 10.263 m /s2.