Answer: 0.19s and 14.6s
Step-by-step explanation:
The distance of the ball at any time is given by the function s(t). For now, you can ignore all the details about the height of the man and the intial velocity of the ball. We only care about what time 't' will the ball be 14 feet off the ground. In math terms, they are asking when:
[tex]s(t)=14[/tex]
Where:
[tex]s(t)=-2.7t^2+40t+6.5=14[/tex]
Subtract 14 on both sides:
[tex]s(t)=-2.7t^2+40t-7.5=0[/tex]
Use the quadratic formula to solve for the variable 't'. This will give the possible time values where the ball will be 14 feet off the ground:
[tex]t=-b+\sqrt{b^2-4ac} /2a[/tex]
[tex]t=-b-\sqrt{b^2-4ac} /2a[/tex]
[tex]a=-2.7[/tex]
[tex]b=40[/tex]
[tex]c=-7.5[/tex]
Therefore:
[tex]t=(-40)+\sqrt{40^2-4(-2.7)(-7.5)} /2(-2.7)=0.19s[/tex]
[tex]t=-40-\sqrt{40^2-4(-2.7)(-7.5)} /2(-2.7)=14.6s[/tex]
This is saying the ball almost immediately hits 14 feet given the intial velocity, then it reaches its peak and falls back down to 14 feet again, where it will eventually hit the ground again
