Ahot-air balloon is rising straight up with a speed of 1.30 m/s. a ballast bag is released from rest relative to the balloon when it is 5.36 m above the ground. how much time elapses before the ballast bag hits the ground?

The time before the ballast hits the ground is 1.186918s

Explanation:

We assume that the upward direction is positive.

The bag released from the ballon is at rest. The velocity of the bag equals the velocity of the balloon.

The balloon has a velocity (V0y) = 3m/s

When the bag is released, it has a height of 5.36m

To touch the ground it makes a displacement of y = - 5.36m

The motion of the bag can be described as followed:

y = V0t + 1/2 * at²

t² + (2V0/a)*t - (2y/a) = 0

We can solve this equation for t:

t= ((-2V0/a) ± √((2V0/a)² + 4*(2y/a))) /2

t = -(V0/a) ± √((V0/a)² + (2y/a))

In this equation we can plug the given values ( V0 = 1.3 m/s ; y = 5.36m ; g =-9.8 m/s²)

t = (1.3/-9.8)±√((1.3/-9.8)²+(2*(-5.36)/-9.8))

t = 0.132653 ± 1.054266

t= 1.186918

The time before the ballast hits the ground is 1.186918s

answer: at 24 february, 2014 the population gets twice of its initial population.

explanation:

since we have given that

the population for alpha city, t years after january 1, 2004 is given as

first we find out the initial population, i.e. t=0,

so, our quadratic equation becomes,

according to question, we have said that if the population has twice its initial population, then it becomes

so, time taken to reach this above mentioned population is given by

now, we know that time can't be negative so, we take t=9.149 years.

hence, at 24 february, 2014 the population gets twice of its initial population.