Question

a ballon is ascending at a rate of 9m/s at a height of 80m above the ground now a packet is dropped from the ballon it will reach the ground in what time ?​

Answer 1

When packet is dropped from balloon, it has same velocity as that of balloon.

Accelration due to gravity (g) = 10 m/s²

(From figure attached)

For journey 1 - 2;

Initial velocity (u) = 9 m/s

Final velocity (v) = 0 m/s

Accelration (a) = -g m/s²

By using $\sf 3^{rd}$ equation of motion:

$\bf {v}^{2} = {u}^{2} + 2as$

$\rm \implies {0}^{2} = {9}^{2} + 2( - g) \times h \\ \\ \rm \implies {0}^{2} = 81 - 20 h \\ \\ \rm \implies 20h = 81 \\ \\ \rm \implies h = \dfrac{81}{20} \\ \\ \rm \implies h = 4.05 \: m$

By using $\sf 1^{rd}$ equation of motion:

$\bf v = u + at$

$\rm \implies v = u + at_{12} \\ \\ \rm \implies 0 = 9 - 10t_{12} \\ \\ \rm \implies 10t_{12} = 9 \\ \\ \rm \implies t_{12} = \dfrac{9}{10} \\ \\ \rm \implies t_{12} = 0.9 \: s$

For journey 2 - 3;

Total/Maximum height $\sf (H_{max})$ = 80 + 4.05 m = 84.05 m

Initial velocity (u) = 0 m/s

a = g m/s²

By using $\sf 2^{nd}$ equation of motion:

$\bf s = ut + \dfrac{1}{2} a {t}^{2}$

$\rm \implies H_{max} = ut_{23} + \dfrac{1}{2} a{t_{23}}^{2} \\ \\ \rm \implies 84.05 = 0 \times t_{23} + \dfrac{1}{2} \times 10{t_{23}}^{2} \\ \\ \rm \implies 84.05 = 0 + 5{t_{23}}^{2} \\ \\ \rm \implies {t_{23}}^{2} = \frac{84.05}{5} \\ \\ \rm \implies {t_{23}}^{2} = 16.81 \\ \\ \rm \implies t_{23} = \sqrt{16.81} \\ \\ \rm \implies t_{23} = 4.1 \: s$

Time taken to reach ground (T) = $\rm t_{12} + t_{23}$

= 0.9 + 4.1

= 5 s

$\therefore$ $\boxed{\mathfrak{Time \ by \ which \ packet \ reach \ ground \ (T) = 5 \ s}}$