Question

A balloon starts rising upwards with constant acceleration $15{\text{m/s}}^2$ and after $10\text{s}$, a packet is dropped from it. The packet reaches the ground after $t_0\text{sec}$. Determine the value of $t_0$ (in sec). Take $g=10{\text{m/s}}^2$ and $\sqrt{15}=3.87$.

• A. $21.4$
• B. $34.33$
• C. $25.5$
• D. $30.7$

Answer 1

The correct option is B $34.33$

Considering vertically upward direction as $+vey-$axis and motion to start at $t=0$

For balloon, $a_y=+15{\text{m/s}}^2$
Displacement of balloon after time $t=10\text{s}$ is given as:
$S=u_{y}t+\frac{1}{2}{a}_y{t}^2$
$\therefore S=0+\frac{1}{2}\times 15\times {10}^2=750\text{m}$
For balloon at $t=10\text{s}$;
$v_y=u_y+a_{y}t=0+15\times 10=150\text{m/s}$

At $t=10\text{s}$, packet's initial velocity will be equal to velocity of balloon at that instant.
$\Rightarrow$ After time $t=t_0\text{s}$, packet hits ground, so its displacement is:
$S_y=-750\text{m}$
$S_y=u_{y}t+\frac{1}{2}{a}_y{t}^2....\left(i\right)$

$t=t_0\text{s},a_y=-g{\text{m/s}}^2,u_y=+150\text{m/s}$
Putting in Eq. $\left(i\right)$, we get:
$\Rightarrow -750=150t_0-\frac{1}{2}\times 10\times t_0^2$
$t_0^2-30t_0-150=\mathrm{0......}\left(ii\right)$

From Eq.$\left(ii\right)$:

$t=\frac{30\pm \sqrt{1500}}{2}$
$\therefore t=15+5\sqrt{15}=3.87(3.87+5)=34.33\text{s}$