A balloon starts rising upward with a constant acceleration ‘a’ and after time t0 second, a packet is dropped from it which reaches the ground after t second. The value of t is

u=at0;h=12at02 for packeth=12gt2−ut12at02=12gt2−at0t12gt2−at0t−12at02=0Solving the quadrant equation, we gett=at021+1+ga

A balloon starts rising upward with a constant acceleration ‘a’ and after time t0 second, a packet is dropped from it which reaches the ground after t second. The value of t is

u=at0;h=12at02 for packeth=12gt2−ut12at02=12gt2−at0t12gt2−at0t−12at02=0Solving the quadrant equation, we gett=at021+1+ga

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A balloon starts rising upward with a constant acceleration ‘a’ and after time t₀ second, a packet is dropped from it which reaches the ground after t second. The value of t is

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$\frac{a t_{0}}{2} [1 + \sqrt{1 + \frac{g}{a}}]$

$a t_{0} [1 + \sqrt{1 + \frac{g}{a}}]$

$\frac{3}{2} a t_{0} [1 + \sqrt{1 + \frac{g}{a}}]$

$\frac{2}{3} a t_{0} [1 + \sqrt{1 + \frac{g}{a}}]$

answer is A.

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$u = a t_{0}; h = \frac{1}{2} a t_{0}^{2}$ for packet

$\begin{array}{l} h = \frac{1}{2} g t^{2} - u t \\ \frac{1}{2} a t_{0}^{2} = \frac{1}{2} g t^{2} - a t_{0} t (\frac{1}{2} g) t^{2} - (a t_{0}) t - \frac{1}{2} a t_{0}^{2} = 0 \end{array}$

Solving the quadrant equation, we get

$t = \frac{a t_{0}}{2} [1 + \sqrt{1 + \frac{g}{a}}]$

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A balloon starts rising upward with a constant acceleration ‘a’ and after time t0 second, a packet is dropped from it which reaches the ground after t second. The value of t is

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A balloon starts rising upward with a constant acceleration ‘a’ and after time t₀ second, a packet is dropped from it which reaches the ground after t second. The value of t is