Answer:
It will take 6 seconds for the ball to reach the ground.
Velocity: [tex]-192\frac{\text{m}}{\text{s}}[/tex].
Step-by-step explanation:
We have been given that a ball dropped from the top of a building has a height of [tex]s =576-16t^2[/tex] meters after t seconds.
The ball will hit the ground, when height will be 0 meters, so we will equate height with 0 as:
[tex]576-16t^2=0[/tex]
Let us solve for t.
[tex]576-576-16t^2=-576[/tex]
[tex]-16t^2=-576[/tex]
[tex]\frac{-16t^2}{-16}=\frac{-576}{-16}[/tex]
[tex]t^2=36[/tex]
Taking positive square root of both sides, we will get:
[tex]\sqrt{t^2}=\sqrt{36}[/tex]
[tex]t=6[/tex]
Therefore, it will take 6 seconds for the ball to reach the ground.
To find the ball's velocity at [tex]t=6[/tex], we will take the derivative of position function and evaluate derivative at [tex]t=6[/tex].
[tex]s=576-16t^2[/tex]
[tex]s'=\frac{d}{dt}(576)-\frac{d}{dt}(16t^2)[/tex]
[tex]s'=0-32t[/tex]
[tex]s'(6)=-32(6)[/tex]
[tex]s'(6)=-192[/tex]
Therefore, the ball's velocity at the moment of impact would be [tex]-192\frac{\text{m}}{\text{s}}[/tex].
