Chapter 6, Problem 88A

To determine

The velocity of the balloon so that Sh catches the bag.

Answer to Problem 88A

The velocity of the balloon so that Sh catches the bag is $v_{\text{balloon}}=8.47\text{m/s}$ .

Explanation of Solution

Given:

The height of the tall building is $\Delta s=60\text{m}$ .

The distance of the hot air balloon from the base is $\Delta x=20\text{m}$ .

The speed with which Cr tosses the jewel bag is $v_{\text{jewel bag}}=7.3\text{m/s}$ .

Formula used:

The relation between the distance $\left(s\right)$ covered by an object, velocity $\left(v\right)$ of the object, and the time taken $\left(t\right)$ is,

  $d=vt$

The equation of motion is given as,

  $s=ut+\frac{1}{2}a{t}^2$

Calculation:

Consider the bag takes $t$ time to fall from the building and travel the horizontal distance $\Delta x$ reach the hands of Sh.

The time $\left(t\right)$ is

  $\begin{array}{l}\Delta x=(v_{\text{jewel bag}})t\\ t=\left(\frac{\Delta x}{v_{\text{jewel bag}}}\right)\end{array}$

Consider Cr tosses the bag from the top of the building.

Consider the initial velocity of the bag is $\left(u_{\text{jewel bag}}\right)$ .

Consider the value of acceleration due to gravity as $g=9.80{\text{m/s}}^2$ .

The downward vertical displacement $\left(\Delta s_1\right)$ covered by the bag is

  $\Delta s_1=(u_{\text{jewel bag}})t+\frac{1}{2}g{t}^{\text{2}}$

The initial velocity of the bag is zero as it starts from rest position.

  $\begin{array}{l}\Delta s_1=(0)t+\frac{1}{2}g{t}^{\text{2}}\\ \Delta s_1=\frac{1}{2}g{t}^{\text{2}}\end{array}$

Substitute $\left(\frac{\Delta x}{v_{\text{jewel bag}}}\right)$ for $t$ .

  $\Delta s_1=\frac{1}{2}g{\left(\frac{\Delta x}{v_{\text{jewel bag}}}\right)}^{\text{2}}$

Consider the balloon is moving upward with the velocity is $v_{\text{balloon}}$ .

The upward vertical distance $\left(\Delta s_2\right)$ covered by the balloon is

  $\begin{array}{l}\Delta s_2=(v_{\text{balloon}})t\\ \Delta s_2=(v_{\text{balloon}})\left(\frac{\Delta x}{v_{\text{jewel bag}}}\right)\end{array}$

The velocity of the bag so that Shifty catches the bag, using the relation is

  $\begin{array}{l}\Delta s=\Delta s_1+\Delta s_2\\ \Delta s=\frac{1}{2}g{\left(\frac{\Delta x}{v_{\text{jewel bag}}}\right)}^{\text{2}}+(v_{\text{balloon}})\left(\frac{\Delta x}{v_{\text{jewel bag}}}\right)\\ (v_{\text{balloon}})\left(\frac{\Delta x}{v_{\text{bag}}}\right)=\Delta s-\frac{1}{2}g{\left(\frac{\Delta x}{v_{\text{jewel bag}}}\right)}^{\text{2}}\\ v_{\text{balloon}}=\left\{\Delta s-\frac{1}{2}g{\left(\frac{\Delta x}{v_{\text{jewel bag}}}\right)}^{\text{2}}\right\}\left(\frac{v_{\text{jewel bag}}}{\Delta x}\right)\end{array}$

Substitute the values of velocity of the balloon, and corresponding distances in the above expression.

  $\begin{array}{l}{v}_{\text{balloon}}=\left\{60-\frac{1}{2}\times 9.80{\left(\frac{20}{7.3}\right)}^{\text{2}}\right\}\left(\frac{7.3}{20}\right)\\ v_{\text{balloon}}=23.22\times 0.365\\ v_{\text{balloon}}=8.47\text{m/s}\end{array}$

Conclusion:

Therefore, the velocity of the balloon so that Sh catches the bag is $v_{\text{balloon}}=8.47\text{m/s}$ .