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Sagot :
The function
[tex]h(t)=-4.9t^2+30.5t+8.8[/tex]Gives us the position of the cannonball measured from the ground at a time t.
When h(t)=0, the cannonball will strike the ground. Therefore,
[tex]\begin{gathered} h(t)=0 \\ \Rightarrow-4.9t^2+30.5t+8.8=0 \end{gathered}[/tex]Solving for t using the quadratic equation formula
[tex]\Rightarrow t=\frac{-30.5\pm\sqrt[]{30.5^2-4(-4.9)(8.8)}}{2\cdot-4.9}=\frac{-30.5\pm33.2074}{-9.8}[/tex]Then,
[tex]\begin{gathered} \Rightarrow t_1=\frac{-30.5+33.2074}{-9.8}=-0.2762\ldots \\ t_2=\frac{-30.5-33.2074}{-9.8}=6.50075\ldots \end{gathered}[/tex]It makes no sense that the cannonball reaches the ground before we shoot it; therefore, t_1 cannot be the answer. The only remaining possibility is t=6.5seconds
The answer is 6.5 seconds.
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