Given the model of the projectile as
[tex]h(t)=-16t^2+91t+3[/tex]
Explanation
Number 1: At height 21, we will have
[tex]\begin{gathered} -16t^2+91t+3=21 \\ -16t^2+91t-18=0 \\ solving\text{ with quadratic formula} \\ t_{1,\:2}=\frac{-91\pm \sqrt{91^2-4\left(-16\right)\left(-18\right)}}{2\left(-16\right)} \\ t_{1,\:2}=\frac{-91\pm \sqrt{7129}}{2\left(-16\right)} \\ \mathrm{The\:solutions\:to\:the\:quadratic\:equation\:are:} \\ t=\frac{91-\sqrt{7129}}{32},\:t=\frac{91+\sqrt{7129}}{32} \\ t=0.21,\:t=5.5 \end{gathered}[/tex]
Answer
t=0.21 seconds
Number 2:
To find the time the projectile will take to touch the ground, we will equate the function to zero.
[tex]\begin{gathered} -16t^2+91t+3=0 \\ solving\text{ with quadratic formula} \\ t_{1,\:2}=\frac{-91\pm \sqrt{91^2-4\left(-16\right)\cdot \:3}}{2\left(-16\right)} \\ t_{1,\:2}=\frac{-91\pm \sqrt{8473}}{2\left(-16\right)} \\ \mathrm{The\:solutions\:to\:the\:quadratic\:equation\:are:} \\ t=-\frac{-91+\sqrt{8473}}{32},\:t=\frac{91+\sqrt{8473}}{32} \\ t=-0.03277,\:t=5.720 \end{gathered}[/tex]
Anwetr t=5.7 seconds
