Answer:
The ball will be 84 feet above the ground 1.125 seconds and 4.5 seconds after launch.
Step-by-step explanation:
Statement is incorrect. Correct form is presented below:
The height [tex]h(t)[/tex] of an ball that is thrown straight upward from an initial position 3 feet off the ground with initial velocity of 90 feet per second is given by equation [tex]h(t) = 3 +90\cdot t -16\cdot t^{2}[/tex], where [tex]t[/tex] is time in seconds. After how many seconds will the ball be 84 feet above the ground.
We equalize the kinematic formula to 84 feet and solve the resulting second-order polynomial by Quadratic Formula to determine the instants associated with such height:
[tex]3+90\cdot t -16\cdot t^{2} = 84[/tex]
[tex]16\cdot t^{2}-90\cdot t +81 = 0[/tex] (1)
By Quadratic Formula:
[tex]t_{1,2} = \frac{90\pm \sqrt{(-90)^{2}-4\cdot (16)\cdot (81)}}{2\cdot (16)}[/tex]
[tex]t_{1} = 4.5\,s[/tex], [tex]t_{2} = 1.125\,s[/tex]
The ball will be 84 feet above the ground 1.125 seconds and 4.5 seconds after launch.
