Question

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

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Accepted Answer

$t = 0$ is the time when the balloon started rising up. At $t = t_{0}$ , 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: $s = u t - \frac{1}{2} g t^{2}$
$\Rightarrow - (\frac{1}{2} a t_{0}^{2}) = (a t_{0}) t - \frac{1}{2} g t^{2} \Rightarrow g t^{2} - 2 a t_{0} t - a t_{0}^{2} = 0$
Solving the above quadratic equation, we get $t = \frac{a t_{0}}{g} [1 + \sqrt{1 + \frac{g}{a}}]$ , because negative root o the expression is not passible as time can not be negative.

B M Sharma Solutions for Chapter: Kinematics I, Exercise 33: ILLUSTRATION 4.31

B M Sharma Physics Solutions for Exercise - B M Sharma Solutions for Chapter: Kinematics I, Exercise 33: ILLUSTRATION 4.31

Questions from B M Sharma Solutions for Chapter: Kinematics I, Exercise 33: ILLUSTRATION 4.31 with Hints & Solutions

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