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 2-34 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 breakpoint does it rise? (b) How high is the break-free point above the ground?

To determine how high above the breakpoint the balloon will rise and the height of the breakpoint from the ground, the kinematics equation for the distance or height can be used.

The initial velocity of balloon, $v_0$, is equal to 19.6 m/s

Time the balloon keeps ascending after the breakpoint$t_1=2s$.

Gravitational Acceleration is $a=g=-9.8\mathrm{m}/{\mathrm{s}}^2$. It is assumed to be negative as it is acting in the downward direction and upward direction is considered as positive.

Time required to reach ground $t_2=8.00s‐2.00s=6s$,

The formula for the height is,

role="math" localid="1656152167610" $h=v_{0}t+\frac{1}{2}a{t}^2$ (i)

Substitute the given information in equation (i) and calculate the height.

$$\begin{gathered} h_1=v_0{t}_1+\frac{1}{2}a{t_1}^2 \\ =19.6m/s\times 2s+\frac{1}{2}\left(-9.8m/s^2\right)\times {\left(2s\right)}^2 \\ =19.6\mathrm{m}\approx 20\mathrm{m} \end{gathered}$$

Therefore, the height above the breakpoint of the balloon is$20\mathrm{m}$.

Putting the given values in the kinematic equation (i),

$$\begin{gathered} h_2=v_0{t}_2+\frac{1}{2}a{t_2}^2 \\ =19.6\frac{m}{s}\times 6s+\left(\frac{1}{2}-\frac{9.8}{s^2}\times 6s^2\right) \\ =58.8\mathrm{m} \\ =59\mathrm{m} \end{gathered}$$

Therefore, the height of the breakpoint from the ground is$59\mathrm{m}$.