Question

A ballon is ascending at the rate of 4.9m/s.A pocket is dropped from the balloon when situated at a height of 24.5m.How long does it take the packet to reach the ground?What is it's final velocity?

Answer 1

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 2

Initially the packet was ascending up with the balloon.

Initially the packet was ascending up with the balloon.Taking upward as positive direction;

Initially the packet was ascending up with the balloon.Taking upward as positive direction;initial velocity, u = 4.9 m/s

Initially the packet was ascending up with the balloon.Taking upward as positive direction;initial velocity, u = 4.9 m/sfinal velocity = v m/s

Initially the packet was ascending up with the balloon.Taking upward as positive direction;initial velocity, u = 4.9 m/sfinal velocity = v m/sinitial height, h₁ = 245 m

Initially the packet was ascending up with the balloon.Taking upward as positive direction;initial velocity, u = 4.9 m/sfinal velocity = v m/sinitial height, h₁ = 245 m final height, h₂ = 0

Initially the packet was ascending up with the balloon.Taking upward as positive direction;initial velocity, u = 4.9 m/sfinal velocity = v m/sinitial height, h₁ = 245 m final height, h₂ = 0a = -9.8 m/s²

Initially the packet was ascending up with the balloon.Taking upward as positive direction;initial velocity, u = 4.9 m/sfinal velocity = v m/sinitial height, h₁ = 245 m final height, h₂ = 0a = -9.8 m/s²time taken = t seconds

Initially the packet was ascending up with the balloon.Taking upward as positive direction;initial velocity, u = 4.9 m/sfinal velocity = v m/sinitial height, h₁ = 245 m final height, h₂ = 0a = -9.8 m/s²time taken = t secondss = ut + 0.5at²

Initially the packet was ascending up with the balloon.Taking upward as positive direction;initial velocity, u = 4.9 m/sfinal velocity = v m/sinitial height, h₁ = 245 m final height, h₂ = 0a = -9.8 m/s²time taken = t secondss = ut + 0.5at²⇒ (h₂-h₁) = ut + 0.5at²

Initially the packet was ascending up with the balloon.Taking upward as positive direction;initial velocity, u = 4.9 m/sfinal velocity = v m/sinitial height, h₁ = 245 m final height, h₂ = 0a = -9.8 m/s²time taken = t secondss = ut + 0.5at²⇒ (h₂-h₁) = ut + 0.5at²⇒ 0-245 = 4.9t + 0.5×(-9.8)×t²

Initially the packet was ascending up with the balloon.Taking upward as positive direction;initial velocity, u = 4.9 m/sfinal velocity = v m/sinitial height, h₁ = 245 m final height, h₂ = 0a = -9.8 m/s²time taken = t secondss = ut + 0.5at²⇒ (h₂-h₁) = ut + 0.5at²⇒ 0-245 = 4.9t + 0.5×(-9.8)×t²⇒ -245 = 4.9t - 4.9t²

Initially the packet was ascending up with the balloon.Taking upward as positive direction;initial velocity, u = 4.9 m/sfinal velocity = v m/sinitial height, h₁ = 245 m final height, h₂ = 0a = -9.8 m/s²time taken = t secondss = 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

Initially the packet was ascending up with the balloon.Taking upward as positive direction;initial velocity, u = 4.9 m/sfinal velocity = v m/sinitial height, h₁ = 245 m final height, h₂ = 0a = -9.8 m/s²time taken = t secondss = 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 =0Solving it, we get t = 7.59s

Initially the packet was ascending up with the balloon.Taking upward as positive direction;initial velocity, u = 4.9 m/sfinal velocity = v m/sinitial height, h₁ = 245 m final height, h₂ = 0a = -9.8 m/s²time taken = t secondss = 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 =0Solving it, we get t = 7.59sv = u + at = 4.9 -9.8×7.59 = 4.9 - 74.38 = -69.48 m/s

Initially the packet was ascending up with the balloon.Taking upward as positive direction;initial velocity, u = 4.9 m/sfinal velocity = v m/sinitial height, h₁ = 245 m final height, h₂ = 0a = -9.8 m/s²time taken = t secondss = 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 =0Solving it, we get t = 7.59sv = u + at = 4.9 -9.8×7.59 = 4.9 - 74.38 = -69.48 m/sSo velocity is 69.48 m/s downward