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Byju's Answer

Standard XII

Physics

Dropping Objects

A balloon sta...

Question

A balloon starts rising upwards with constant acceleration a and after time $t_{0}$ second, a packet is dropped from it which it reaches the ground after t seconds of dropping. Determine the value of t

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Solution

Analysis of situation: t = 0 is the time when the balloon started rising up. At t = 0, when the packet is dropped, the balloon is moving up with velocity v = 0 + $a t_{0}$ = $a t_{0}$ . Hence, initial velocity of the packet will be $v_{0} = a t_{0}$ (upward). As the balloon has started rising upwards with constant acceleration a, so after $t_{0}$ seconds, its height from the ground is $y_{0} = \frac{1}{2} a t_{0}^{2}$ .
For packet: s = ut - $\frac{1}{2} g t^{2}$
$\Rightarrow$ $- \frac{1}{2} a t_{0}^{2}$ = $a t_{0} t - \frac{1}{2} g t^{2}$ $\Rightarrow$ $g t^{2} - 2 a t_{0} t - a t_{0}^{2} = 0$
Solving the quadratic equation, we get t = $\frac{a t_{0}}{g} [1 + \sqrt{1 + \frac{g}{a}}]$

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A balloon starts rising upwards with constant acceleration a and after time t0 second, a packet is dropped from it which it reaches the ground after t seconds of dropping. Determine the value of t

A balloon starts rising upwards with constant acceleration a and after time $t_{0}$ second, a packet is dropped from it which it reaches the ground after t seconds of dropping. Determine the value of t