Will mark as BRAINLIEST A balloon is ascending at the rate of 4.9 m/s. A packet is dropped from from the balloon when situated at a height of 100m.
How long does it take the packet to reach the ground ?
What is it's final velocity ?

Explanation:

Initially the packet was ascending up with the balloon.

Taking upward as positive direction;

initial velocity, u = 4.9 m/s

final velocity = v m/s

initial height, h₁ = 100 m  

final height, h₂ = 0

a = -9.8 m/s²

time taken = t seconds

Considering the second equation of motion.

S = ut + 1/2at^2

s = ut + 0.5at²

(h₂-h₁) = ut + 0.5at²

0- 100 = 4.9t + 0.5×(-9.8)×t²

-100 = 4.9t - 4.9t²

4.9t² -4.9t -  100 =0

t² -t - 20.41 = 0

Using : b±√(b²-4ac))/(2a)....  1

Where a =1, b = -1 c =-20.408

Substituting the values into equation 1

Solving it, we get  t = 5.045s

v = u + at = 4.9 -9.8×5.045 = 4.9 - 49.44 = -44.54 m/s

The final velocity is -44.54 m/s, and it took  5.045s to reach the ground.

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s

i think its c if i'm wrong