Given:
[tex]h(t)=-16t^2+95t+3[/tex]
Find-:
How long will it take for the ball to hit the ground?
Explanation-:
To hit the ground height is zero.
[tex]h(t)=0[/tex][tex]\begin{gathered} h(t)=-16t^2+95t+3 \\ \\ -16t^2+95t+3=0 \end{gathered}[/tex]
So, the time is:
[tex]\begin{gathered} -16t^2+95t+3=0 \\ \\ \end{gathered}[/tex]
Use quadratic formula:
[tex]\begin{gathered} ax^2+bx+c=0 \\ \\ x=\frac{-b\pm\sqrt{b^2-4ac}}{2a} \end{gathered}[/tex]
So, the value of "t" is:
[tex]\begin{gathered} -16t^2+95t+3=0 \\ \\ t=\frac{-95\pm\sqrt{95^2-4(-16)(3)}}{2(-16)} \\ \\ t=\frac{-95\pm\sqrt{9025+192}}{-32} \\ \\ t=\frac{-95\pm96.005}{-32} \\ \\ t=5.969,t=-0.03 \end{gathered}[/tex]
So, after 5.969 second ball to hit the ground.
