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An arrow is shot vertically upward from a platform 46 ft high at a rate of 183 ft/sec When will the arrow hit the groundUse the formula h=-16t^ 2 +v 0 f+h 0 . Round your answer to the nearest )

Respuesta :

Here

[tex]v_o=183\text{ ft/sec}[/tex][tex]h_o=46\text{ ft}[/tex]

When the arrow hit the ground then h=0 so

[tex]0=16t^2+183t+46[/tex]

[tex]\Rightarrow16t^2+183t+46=0[/tex]

Solving the equation we have

[tex]t=\frac{-183\pm\sqrt[]{183^2+2944}}{32}=\frac{-183\pm\sqrt[]{33489+2944}}{32}=\frac{-183\pm\sqrt[]{36433}}{32}=\frac{-183\pm190.8743}{32}[/tex][tex]t=\frac{-183+190.8743}{32}=0.2461\text{ }[/tex]

or

[tex]t=\frac{-183-190.8743}{32}=\frac{-373.8743}{32}=-11.685[/tex]

Since time can not be negative so the time required when the arrow hit the ground will be

[tex]t=0.2461\text{ sec}[/tex]

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