Free-Fall Acceleration — Question 3

Question 3

A released package inherits the carrier velocity

A balloon rises vertically at a constant ground speed of $10\ \mathrm{m/s}$. At height $40\ \mathrm{m}$ it releases a small package without a push. Thereafter the balloon continues at the same speed, while the package experiences free fall with $g=10\ \mathrm{m/s^2}$ and negligible air resistance. Set $t=0$ at release and use upward-positive ground height.

1. Identify the package’s initial ground velocity and write both height functions. Does the package immediately descend relative to the ground?

2. Find the package’s highest point and ground-contact time, including its velocity just before contact.

3. Derive the package’s position relative to the balloon. Evaluate the separation at the package’s highest point and just before ground contact.

4. Explain how the package can rise relative to the ground while moving downward relative to the balloon. What contact time would the incorrect assumption of zero initial ground velocity predict?

Question 3 - Solution

A released package inherits the carrier velocity

(a) No release push means no instantaneous velocity change.

The initial package velocity is $\boxed{+10\ \mathrm{m/s}}$, inherited from the balloon. With numerical seconds and metres, $$\boxed{y_P=40+10t-5t^2},\qquad\boxed{y_B=40+10t}.$$ The package velocity is $v_P=10-10t$, positive for $0<t<1\ \mathrm{s}$. Release does not make it instantly move downward relative to the ground.

(b) Locate the turn and the first contact separately.

At $\boxed{t=1\ \mathrm{s}}$, $v_P=0$ and $\boxed{y_{\max}=45\ \mathrm{m}}$. Ground contact requires $t^2-2t-8=0$, giving roots 4 s and $-2$ s. The admissible event is $$\boxed{T=4\ \mathrm{s}},\qquad\boxed{v_P(T^-)=10-40=-30\ \mathrm{m/s}}.$$ The negative root precedes release and is excluded.

(c) Subtract positions measured at the same time.

$$\boxed{y_P-y_B=-\tfrac12gt^2=-5t^2\quad\text{(metres)}}.$$ At 1 s, the package is 5 m below the balloon: their heights are 45 m and 50 m. At 4 s, the package is at ground level and the balloon at 80 m, a separation of $\boxed{80\ \mathrm{m}}$.

(d) Relative velocity and ground velocity answer different questions.

$$0<t<1\ \mathrm{s}:\quad v_P=10-10t>0,\qquad v_P-v_B=-10t<0\quad\text{(SI)}.$$

Relative velocity is $v_P-v_B=-gt<0$ for every $t>0$. During the first second both objects rise, but the package rises more slowly and falls farther behind. There is no contradiction.

Incorrectly setting initial ground velocity to zero would give $40-5t^2=0$, or $\boxed{T_{\mathrm{wrong}}=\sqrt8\approx2.828\ \mathrm{s}}$. That describes release from rest relative to the ground, not release from the rising balloon.