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:

Given parameters

initial velocity u = 4.9m/s

height of drop H = 100m

acceleration due to gravity g = -9.8m/s²(since the body is ascending i.e moving against gravity)

Required

time taken by the packet to reach the ground 't'

Using the equation of motion

S = ut+1/2gt²

0-100 = 4.9t+1/2(-9.8)t²

-100 = 4.9t-1/2(9.8)t²

-100 = 4.9t-4.9t²

4.9t-4.9t² +100 = 0

4.9t²-4.9t² -100 = 0

Multiplying through by 10

49t²-49t-1000 = 0

Using the general equation t = (-b±√b²-4ac)/2a

t = -(-49)±√(-49)²-4(49)(-1000)/2(49)

t = 49±√(2401+196000)/2(49)

t = 49±√(198401)/98

t = 49±√(198401)/98

t = 49±449.42/98

t = (49+449.42)/98

t = 494.42/98

t = 5.04secs

Hence, it took the packet 5.04secs to reach the ground

The final velocity is calculated using the formula v = u + at

v = u-gt

v = 4.9-(9.8)(5.04)

v = 4.9-49.392

v = -44.492m/s

This shows that the final velocity is 44.49m/s(moving downwards)

option number c is the right answer.

find the surface area of the sphere to the nearest square unit. use a calculator.

a. 254 in.2

b. 64 in.2

c. 127 in.2

d. 1,018 in.2