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.
= 509.5 m. vA
= 98.1 m/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,
= VA + a tB.
Since 10 s is less than the time of jam, 13 s, our assumption above is justified.
- VA2 = 2 a sAB
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
= vB tBC + 0.5 a tBC2.
The velocity at C
is given by
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.
- vC2 = 2 (anet)(sC)
Acceleration due to
gravity is always present and is acting downwards at any instant of time.