As a runaway scientific balloon ascends at 19.6 m/s, one of its instrument packages breaks free of a harness and free-falls. Figure below gives the vertical velocity of the package versus time, from before it breaks free to when it reaches the ground. (a) What maximum height above the break-free point does it rise? (b) How high is the break-free point above the ground?
Take g = 9.8 m/s2
above break free point 19.6 m,
height of break free point 58.8
Till t = 2 sec the velocity of package is same as that of balloons i.e., 19.6 m/s
As soon as the parcel leaves the balloon its velocity is + 19.6 m/s (upward). Its acceleration is - 9.8 m/s2(downwards)
Now the parcel will keep going up till its velocity becomes 0.
So v = 0 u = +19.6 a = - 9.8 m/s2
v2−u2=2as
(19.6)2= 2(-9.8) s
9.8×42=s=19.6m
So, parcel goes 19.6 m above the break free point.
The parcel broke free at t = 20 sec. with v= + 19.6
The parcel hits ground at t = 8 sec
So time of flight = 8 - 2 = 6 sec
u = 19.6 t = 6 sec a = - 9.8 m/s2
s=19.6×6+12+(−9.8)(6)2
s=−58.8m
Negative sign says that ball was displaced by 58.8m downwards from the breakaway point.