A basketball player shoots toward a basket 4.9 m away and 3.0 m above the floor. If the ball is released 1.8 m above the floor at an angle of 60 o above the horizontal, what must the initial speed be if it were to go through the basket

  v₀ = 6.64 m / s

Explanation:

This is a projectile throwing exercise

          x = v₀ₓ t

          y = y₀ + v_{oy} t - ½ g t²

In this case they indicate that y₀ = 1.8 m and the point of the basket is x=4.9m y = 3.0 m

the time to reach the basket is

        t = x / v₀ₓ

we substitute

        y- y₀ =

        y - y₀ = x tan θ -

we substitute the values

        3 -1.8 = 3.0 tan 60 -

        1.2 = 5.196 - 176.4 1 / v₀²

        176.4 1 / v₀² = 3.996

        v₀ =

        v₀ = 6.64 m / 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.