Answer:
Step-by-step explanation:
In your formula, the -16t-squared term is representative of the acceleration due to gravity in feet per second squared. The next term v0 is the initial upwards velocity and h0 is the initial height from which the object was either dropped or launched upwards (and then gravity takes over and pulls it back down as gravity does!). If Michael is dropping the ball, it is in free fall and because he dropped it as opposed to throwing it up into the air creating parabolic motion, there is no upwards velocity at all. In other words, v0 = 0. The height from which Michael dropped the ball is 20 feet, so h0 = 20 and our quadratic is
[tex]s(t)=-16t^2+20[/tex]
where s(t) is the position of the ball after a certain amount of time.
If we are asked when the ball will hit the ground, we know that ON the ground, the height of the ball is 0, right? Sub in 0 for s(t) and solve for t, the time it takes the ball to hit the ground. This is also the same exact thing as "how long was the ball in the air?"
[tex]0=-16t^2+20[/tex] and, factoring out whatever we can (which is very little):
[tex]0=-4(4t^2-5)[/tex]
By the Zero Product Property, either -4 = 0 (which OF COURSE IT DOESN'T!!) or
[tex]4t^2-5=0[/tex]
Solving for t gives us
[tex]4t^2=5[/tex] and
[tex]t^2=+/-\frac{5}{4}[/tex] and
[tex]t=+/-\frac{\sqrt{5} }{2}[/tex] which simplifies in decimals to either -1.12 seconds or 1.12 seconds. We all know that time will never be negative, so the time it takes for the ball to hit the ground is 1.12 seconds.
Learn this stuff as best as you can because it is the foundation for derivatives in calculus and two-dimensional motion in physics!
