Question

The answer is 1.09 s, I just would like to see how to solve it

A hot-air balloon is rising upward with a constant speed of 2.60 m/s. When the balloon is 3.02 m above the ground, the balloonist accidentally drops a compass over the side of the balloon. How much time elapses before the compass hits the ground? 1.09 s (in s)

Answer 1

Here balloon is moving upward so upward velocity is positive and compass is moving downward so it's velocity would be negative.

ли Given u=-2.60 mis h = 3.02 m here g = +9.8 m/s² , because balloon is moving up ward & Compass moves downward. so from equation of motion under gravity h= ut + 4 gt² balloon 3.02 = -2.60 t t Į (9.8) t? compass 3.02m! = 4.9t? - 2.60 € - 3.02 = 0 | Now from quadratic formula t = + 2.60 $ 162.60)2- 4(4.9)( -3.02) 2(4.9) = +2.60 f.J 6.76 + 59.92 9.8 = +2.60 + 9.8 8:12 - = log & -0.56 time has no significance. but we know negative - It = 1.09 sec 1