Question

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 $m/s^2$

• A. $h_{max}$above break free point 19.6 m height of break free point 78.4
• B. $h_{max}$ above break free point 78.4 m, height of break free point 19.6
• C. $h_{max}$ above break free point 19.6 m, height of break free point 58.8
• D. $h_{max}$ above break free point 58.8 m, height of break free point 9.8

Answer 1

The correct option is C

$h_{max}$ 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 $m/s^2$(downwards)

Now the parcel will keep going up till its velocity becomes 0.

So v = 0 u = +19.6 a = - 9.8 $m/s^2$

$v^2-u^2=2as$

$(19.6{)}^2$= 2(-9.8) s

$\frac{9.8\times 4}{2}=s=19.6m$

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 $m/s^2$

$s=19.6\times 6+\frac{1}{2}+(-9.8)(6{)}^2$

$s=-58.8m$

Negative sign says that ball was displaced by 58.8m downwards from the breakaway point.