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Answer

We know that,

s = ut + 0.5at² ⇒ (h₂-h₁) = ut + 0.5at² ⇒ 0-245 = 4.9t + 0.5×(-9.8)×t² ⇒ -245 = 4.9t - 4.9t² ⇒ 4.9t² -4.9t -245 =0 Solving it, we get  t = 7.59s v = u + at = 4.9 -9.8×7.59 = 4.9 - 74.38 = -69.48 m/s So, velocity is 69.48 m/s downward

Answer

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 = 245 m   final height, h₂ = 0 a = -9.8 m/s² time taken = t seconds s = ut + 0.5at² ⇒ (h₂-h₁) = ut + 0.5at² ⇒ 0-245 = 4.9t + 0.5×(-9.8)×t² ⇒ -245 = 4.9t - 4.9t² ⇒ 4.9t² -4.9t -245 =0 Solving it, we get  t = 7.59s v = u + at = 4.9 -9.8×7.59 = 4.9 - 74.38 = -69.48 m/s So velocity is 69.48 m/s downward  

Answer

First you have to calculate the final velocity.  For a particle thrown from a height, initial velocity is zero as it is seperated from the balloon. Hence,

v2=2gs (u=0)=2x10x245, v=70m/s

v=gt, hence t=70/10=7 s

Answer

Kindly go through the desired result.

 

Answer

initial velocity of the packet , u = 4.9 m/s final velocity = v m/s  height, h = 245 m   g = 9.8 m/s² time taken = t seconds h = ut + 0.5gt²    [ since u is upward it is taken as negative) 245=-4.9t + 0.5(9.8)t² ⇒245 = -4.9t + 0.5×(9.8)×t² ⇒ 245 =- 4.9t + 4.9t² ⇒ 4.9t² -4.9t -245 =0 ⇒ t² -t -50 =0  Solving it, we get  t = 7.59s 0R  t = -6.59s (discarded since time can't be negative)

v = u + gt = -4.9 +9.8(7.59) = -4.9 +74.38 = 69.48 m/s  So final velocity is 69.48 m/s downward  

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