# A balloon is ascending at the rate of 4.9m/s. A pocket is dropped from the balloon when situated at a height of 245m. How long does it take the packet to reach the ground? What is its final velocity? CLASS - XI PHYSICS (Kinematics)

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2014-10-11T21:41:49+05:30

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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