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.