Q: A balloon is going upwards with velocity 12 m/sec. It release a packet when it is at a height of 65 m from the ground. How much time the packet will take to reach the ground if g = 10 m/s^{2}

(a) 5 sec

(b) 6 sec

(c) 7 sec

(d) 8 sec

Ans:(a)

Sol: As packet is to fall to the ground and it is released initially with the velocity of balloon, so

u =-12 m/s,

$ \displaystyle s = ut + \frac{1}{2}at^2 $

$ \displaystyle 65 = -12t + \frac{1}{2}(10)t^2 $

$ \displaystyle 65 = -12t + 5 t^2 $

$\displaystyle 5 t^2 -12 t – 65 = 0 $

$ \displaystyle t = \frac{12 \pm \sqrt{144 – 4(5)(-65)}}{2\times 5} $

$ \displaystyle t = \frac{12\pm \sqrt{144 + 1300}}{10} $

$ \displaystyle t = \frac{12 \pm 38}{10} $

t = 5 sec