Answer:
≈ 6 seconds
Step-by-step explanation:
When it says about hitting the ground, it means it hits the t-axis.
That means we have to solve for t-intercepts to find the solution.
Finding t-intercept, let y = 0.
[tex] \displaystyle \large{0 = - 16 {t}^{2} + 96t + 3}[/tex]
Use the Quadratic Formula.
[tex] \displaystyle \large{t = \frac{ - b \pm \sqrt{ {b}^{2} - 4ac } }{2a} }[/tex]
Substitute in a = -16, b = 96 and c = 3.
[tex] \displaystyle \large{t = \frac{ - 96 \pm \sqrt{ {96}^{2} - 4( - 16)(3)} }{2( - 16)} } \\ \displaystyle \large{t = \frac{ - 96 \pm \sqrt{ 9216 + 192} }{ - 32} } \\ \displaystyle \large{t = \frac{ - 96 \pm \sqrt{ 9408} }{ - 32} } \\ \displaystyle \large{t = \frac{ - 96 \pm 56 \sqrt{3} }{ - 32} } \\ [/tex]
Then simplify to simplest form.
[tex] \displaystyle \large{t = \frac{ - 96 \pm 56 \sqrt{3} }{ - 32} } \\ \displaystyle \large{t = \frac{ 12\pm 7\sqrt{3} }{ 4} } \\ [/tex]
Then use the highest value/root which is 12+7√3 over 4 then input it in a calculator and you will see that the highest root is close to 6.
We use the highest root because thats when y or height = 0 hence if height = 0 then the object finally hits the ground.
