Question

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

Answer 1

Velocity of packet when dropped will be same as velocity of balloon ,but as its acceleration becomes g downward,

velocity of balloon after time t will be

$v=u+a\times 2$

since u=0

$v=a\times 2$

and distance travelled by balloon will be

$s=u\times t+\frac{1}{2}\times a\times t^2$

$s=0+\frac{1}{2}\times a\times 4$

$s=2a$

so velocity of packet will also be same

for packet

$u_{initial}=2a$ $a=-g$ . $time=t$ and $s=-2a$

$s=u\times t+\frac{1}{2}\times a\times t^2$

$-2a=2a\times t+\frac{1}{2}\times -g\times t^2$

solving above quadratic equation

$t=\frac{a+\sqrt{a^2+8ag}}{2g}$

this is the time required for the packet to reach ground.