Question

A balloon is ascending at the rate of 5 m/s at a height of 100 m above the ground when a packet
is dropped from the balloon. After how much time does it reach the ground ? (g = 10 m/s)
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Answer 1

hey here is your answer.

v=v0-gt=5–9.81t

s=s0+v0t-0.5gt^2

s=100+5t-0.5(9.81)t^2

when it reaches the ground s=0

9.81t^2–10t-200=0

t={-(-10)+/-[(-10)^2–4(9.81)(-200)]^0.5}/2(9.81)

={10+/-[100+7848]^0.5}/19.62

={10+/-[7948]^0.5}/19.62={10+/-89.15}/19.62=5.05 sec

only the positive root has meaning.

Therefore it will take about 5.05 seconds to hit the ground.

hope it will help you.

plss brainliest my answer.

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Answer 2

v=v0-gt=5–9.81t

s=s0+v0t-0.5gt^2

s=100+5t-0.5(9.81)t^2

when it reaches the ground s=0

9.81t^2–10t-200=0

t={-(-10)+/-[(-10)^2–4(9.81)(-200)]^0.5}/2(9.81)

={10+/-[100+7848]^0.5}/19.62

={10+/-[7948]^0.5}/19.62={10+/-89.15}/19.62=5.05 sec

Answer:

Explanation: plz brainlist me...