Aprojectile is fired straight upward with an initial veloc- ity of 100 m=s from the top of a building 20 m high and falls to the ground at the base of the building. find (a) its maximum height above the ground; (b) when it passes the top of the building; (c) its total time in the air.

The answer to your question is:

a) h max = 529.7 m

b) t = 20.4 s

c) t = 20.6 s

Explanation:

a) h max = -(vo)² / 2g

              = 100² / 2(9.81)

              = 10000 / 19.62

              = 509.7 m

total height = 509.7 + 20 = 529.7 m

b)  

    h = gt² / 2

    t = √ 2h / g

    t = √ 2(509.7)/9.81

    t = √ 103.91

    t = 10.19 s  

    total time = 2 x t = 2 x 10.19 = 20.4 s

c)

   h = vot + 1/2gt²

   20 = 100t + 1/2(9.91) t²

  4.9t² + 100 t -20 = 0   quadratic equation

t = 0.19 s

Total time = 0.19 + 20.4 = 20.6 s

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.