Answer:
t = 4.6 seconds
Step-by-step explanation:
The ball's height above the ground t seconds later is given by the function is given by :
[tex]h(t)=-16t^2+48t+120[/tex]
We need to find time when the ball be 20 feet above the ground. It means h(t) = 20.
So,
[tex]-16t^2+48t+120=20\\\\-16t^2+48t+100=0[/tex]
The above is a quadratic equation, the solution of this quadratic equation is given by :
[tex]t=\dfrac{-b\pm \sqrt{b^2-4ac} }{2a}\\\\t=\dfrac{-b+ \sqrt{b^2-4ac} }{2a},\dfrac{-b- \sqrt{b^2-4ac} }{2a}\\\\t=\dfrac{-48+ \sqrt{(48)^2-4\times (-16)(120)} }{2\times (-16)} \dfrac{-48-\sqrt{(48)^2-4\times (-16)(120)} }{2\times (-16)}\\\\t=-1.62\ s, 4.6\ s[/tex]
So, at t = 4.6 seconds, the ball will be 20 feet above the ground.
