Question

Q(a) A balloon is ascending at the rate of 14 m/s at a height of 98 m above the ground, when a packet is dropped from it. After how much time and with what velocity does it reach the ground?​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

A balloon is ascending at the rate of 14 m/s at a height of 98 m above the ground, when a packet is dropped from it.

It means, initial velocity of packet is 14 m/s.

Let assume that, it took t seconds to reach the ground.

Height of packet, while it is dropped is 98 m.

So, we have,

Distance, s = - 98 m [ sign convention ] as its ascending vertically.

Acceleration due to gravity, g = - 9.8 m/s^2

Initial velocity, u = 14 m/s

Using, Equation of motion,

$\rm :\longmapsto\: \boxed{ \bf{ \: s = ut + \dfrac{1}{2} g {t}^{2}}}$

On substituting the values, we get

$\rm :\longmapsto\: - 98 = 14t - \dfrac{1}{2} \times 9.8 \times {t}^{2}$

On multiply by - 1, we get

$\rm :\longmapsto\:98 = - 14t + 4.9{t}^{2}$

$\rm :\longmapsto\:980 = - 140t + 49{t}^{2}$

$\rm :\longmapsto\:140 = - 20t + 7{t}^{2}$

$\rm :\longmapsto\:7{t}^{2} - 20t - 140 = 0$

Its a quadratic equation in 't', so using Quadratic Formula,

$\boxed{ \bf{ \: t \: = \: \frac{ - b \: \pm \: \sqrt{ {b}^{2} - 4ac} }{2a}}}$

On Substituting the values, we get

$\rm :\longmapsto\:t = \dfrac{20 \: \pm \: \sqrt{400 + 3920} }{2 \times 7}$

$\rm :\longmapsto\:t = \dfrac{20 \: \pm \: \sqrt{4320} }{14}$

$\rm :\longmapsto\:t = \dfrac{20 \: \pm \: 65.728 }{14}$

$\rm :\longmapsto\:t = \dfrac{20 \: + \: 65.728 }{14} \: \: as \: t \cancel < 0$

$\rm :\longmapsto\:t = \dfrac{ 85.728 }{14} \:$

$\bf\implies \:t = 6.123 \: seconds \:$

Now, To find the velocity by which packet reach the ground.

Let assume that the velocity of packet be v m/s.

Now, we have

Initial velocity of packet, u = 14 m/s

Height of packet, s = 98 m

Acceleration due to gravity, g = 9.8 m/s^2

So, By using equation of motion,

$\rm :\longmapsto\: {v}^{2} - {u}^{2} = 2gs$

On substituting the values, we get

$\rm :\longmapsto\: {v}^{2} - {14}^{2} = 2 \times 9.8 \times 98$

$\rm :\longmapsto\: {v}^{2} - 196 = 1920.8$

$\rm :\longmapsto\: {v}^{2} = 1920.8 + 196$

$\rm :\longmapsto\: {v}^{2} = 2116.8$

$\rm :\longmapsto\: {v} = \sqrt{ 2116.8}$

$\bf :\longmapsto\: {v} = 45.99 \approx \: 46 \: m \: per \: sec$

Answer 2

Let us take the point from where the food packet is dropped as origin of coordinate system, upwards direction as positive and downwards direction as negative. The food packet on being dropped from an ascending balloon, would also acquire the velocity of the balloon ie an upward velocity of 14 m/s. Besides, the food packet would also be acted upon by the acceleration due to gravity a = 9.8 m/s² directed downwards.

We can use the relation;

s = u t + ½ a t²,

where,

u = initial velocity of the food packet = +14 m/s ( + sign because the velocity is directed upwards)

a = acceleration due to gravity = - 9.8 m/s² (- ve sign as the acceleration due to gravity is directed downwards)

s = displacement of the food packet = - 98 m ( -ve sign as ground is downwards from the drop point).

Substituting various values we get,

- 98 = 14 t - ½ × 9.8 × t²

=> - 98 = 14 t - 4.9 t²

=> 4.9 t² -14 t - 98 = 0

Dividing by 0.7 throughout we get

7 t² - 20 t - 140 = 0

Solving for t we get,

t = [( 20 ± 65.73)/14] = - 3.266 s or 6.12 s.

Time t = - 3.26 s is unphysical. It means that the packet reached the ground before being dropped.

So the time t taken by the food packet to reach the ground = 6.12 second after being dropped.

The vocity v with which the food packet reaches the ground can be obtained using the relation:

v = u + a t = 14 - 9.8 × 6.12 = -45.98 m/s.

The minus sign with v mean that the velocity of the food packet is directed downwards.