Question

a balloon is ascending at rate of 5m/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​

Answer 1

Explanation:

Δy=vt+12at2Δy=vt+12at2

−100=5t+12(−9.8)t2−100=5t+12(−9.8)t2

4.9t2−5t−100=04.9t2−5t−100=0

t=5±25−4(4.9)(−100)√2(4.9)t=5±25−4(4.9)(−100)2(4.9)

t=5±1985√9.8t=5±19859.8

t≈5±44.55339.8≈5.056,−4.036t≈5±44.55339.8≈5.056,−4.036

We can throw out the negative answer since time is positive.

Therefore it will take about 5.0565.056 seconds to hit the ground.

1.7K views

View 3 Upvoters

Related Questions (More Answers Below)

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

only the positive root has meaning.