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Sagot :
1) The function given is a quadratic function with negative leading coefficient, so its maximum value can be calculated using the formula for the y value of the vertex of a parabola:
[tex]\begin{gathered} y_v=-\frac{\Delta}{4a} \\ \Delta=b^2-4ac \end{gathered}[/tex]Since the function is give, we can get the values for a, b and c:
[tex]\begin{gathered} y=ax^2+bx+c \\ h(t)=-16t^2+30t+1000 \end{gathered}[/tex]So:
[tex]\begin{gathered} \Delta=30^2-4(-16)(1000)=900+64000=64900 \\ h_{\max }=-\frac{64900}{4(-16)}=-\frac{64900}{-64}=1014.0625 \end{gathered}[/tex]So, the maximum height is 1014.0625 feet.
2) This is also the vertex of the parabola, but now the x value formula:
[tex]x_v=-\frac{b}{2a}[/tex]So:
[tex]t_{\max }=-\frac{30}{2(-16)}=-\frac{30}{-32}=0.9375[/tex]So, it take 0.9375 seconds to reach the maximum.
3) When it hits the ground, h(t) = 0, so this is the same as fiding the zero of the function, which we can do by using the quadratic formula:
[tex]\begin{gathered} t=\frac{-b\pm\sqrt[]{\Delta}}{2a}=\frac{-30\pm\sqrt[]{64900}}{2(-16)}=\frac{-30\pm254.754\ldots}{-32} \\ t_1=\frac{-30+254.754\ldots}{-32}=\frac{224.754\ldots}{-32}=-7.02358\ldots \\ t_2=\frac{-30-254.754\ldots}{-32}=\frac{-284.754\ldots}{-32}=8.89858\ldots \end{gathered}[/tex]One of the results is negative, so it doen's make sence for this situation. So, the only solution is:
[tex]t=8.89858\ldots\approx8.8986[/tex]So, it take approximately 8.8986 seconds for it to hit the gournd.
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