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Sagot :
We will have the following:
***First:
[tex]h=h_0+v_0\cdot t+\frac{1}{2}g\cdot t^2[/tex]Now, we will determine the value for the speed:
[tex]1840=1600+v_0\cdot(4)+\frac{1}{2}(-32.17)\cdot(4)^2\Rightarrow240=4v_0-\frac{25736}{25}[/tex][tex]\Rightarrow\frac{31736}{25}=4v_0\Rightarrow v_0=\frac{7934}{25}\Rightarrow v_0=137.36[/tex]So, the equation for the height of the arrow (h) in feet as a function of the number of seconds t is:
[tex]h=1600+317.36t+\frac{1}{2}gt^2[/tex]Here "g" is the gravitational pull of earth.
***Second:
We will determine how much time it would take for the arrow to hit the ground as follows:
[tex]0=1600+317.36t+\frac{1}{2}(-32.17)t^2\Rightarrow-\frac{3217}{200}t^2+317.36t+1600=0[/tex][tex]\Rightarrow t=\frac{-(317.36)\pm\sqrt[]{(317.36)^2-4(-\frac{3217}{200})(1600)}}{2(-\frac{3217}{200})}\Rightarrow\begin{cases}t\approx-4.163 \\ t\approx23.893\end{cases}[/tex]So, afeter 23.893 seconds the arrow would hit the ground.
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