Question

A balloon is ascending at a rate of 14 m/s at a height of 98 m above the ground when the food packet is dropped from the balloon. after how much time and with what velocity does it reach the ground? Take g = 9.8m/s².

${\boxed{\undeline{\red{instà :- theanuragkumar_}}}}$​

Answer 1

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

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.

Answer 2

Answer:

Given,

Velocity of the balloon = initial velocity of the food packet u=14j^ms−1

Gravity acceleration a=−9.81 j^   m/s2

Distance from ground h=−98 j^ m

Final velocity =v & time taken to reach the ground =t

Apply second kinematic equation

  v2−u2=2ah

 v=2ah+u2

 v=2(−9.81)(−98)+142

 v=−46j^ m/s

Apply first kinematic equation

  v=u+at

 t=av−u=−9.81−46−14=6.11 sec 

At velocity 46m/s  and in time 6.11 sec it will reach to the ground.