Answer:
Therefore the ball strikes the ground 5.98 s after kick.
Step-by-step explanation:
Given that , a ball is kicked in the air. The height of the ball can be calculated using the equation [tex]h(t)=-16t^2+64t+190[/tex] where h is height in feet and t is time in second.
When the ball touches the ground then the height will be zero.
i.e h(t)=0
Therefore
[tex]-16t^2+64t+190=0[/tex]
[tex]\Rightarrow -2(8t^2-32t-95)=0[/tex]
Since -2≠0 then [tex](8t^2-32t-95)=0[/tex]
[tex](8t^2-32t-95)=0[/tex]
[The solution of a quadratic equation ax²+bx+c=0 is
[tex]x=\frac{-b\pm\sqrt{b^2-4ac} }{2a}[/tex] here a=8, b=-32 and c=-95]
[tex]\Rightarrow t =\frac{-(-32)\pm\sqrt{(-32)^2-[4\times8\times(-95)]} }{2\times8}[/tex]
[tex]\Rightarrow t =\frac{32\pm\sqrt{4064} }{16}[/tex]
Therefore t= 5.98 or -1.98
since time can not be negative So , t=5.98 s
Therefore the ball strikes the ground 5.98 s after kick.
