Solution:
Given:
[tex]h=141-21t-16t^2[/tex]To get the time the ball hit the ground, it hits the ground when the height is zero.
Hence,
[tex]\begin{gathered} At\text{ h = 0;} \\ h=141-21t-16t^2 \\ 0=141-21t-16t^2 \\ 141-21t-16t^2=0 \\ 16t^2+21t-141=0 \end{gathered}[/tex]To solve for t, we use the quadratic formula.
[tex]\begin{gathered} x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ \text{where;} \\ a=16,b=21,c=-141 \\ t=\frac{-21\pm\sqrt[]{21^2-(4\times16\times-141)}}{2\times16} \\ t=\frac{-21\pm\sqrt[]{441+9024}}{32} \\ t=\frac{-21\pm\sqrt[]{9465}}{32} \\ t=\frac{-21\pm97.288}{32} \\ t_1=\frac{-21+97.288}{32}=\frac{76.288}{32}=2.384\approx2.38 \\ t_2=\frac{-21-97.288}{32}=\frac{-118.288}{32}=-3.6965\approx-3.70 \end{gathered}[/tex]
Since time can't be a negative value, we pick the positive value of t.
Therefore, to the nearest hundredth, it takes 2.38 seconds for the ball to hit the ground.
