Question

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 )

Answer 1

Here

`v_o=183\text{ ft/sec}`
`h_o=46\text{ ft}`

When the arrow hit the ground then h=0 so

`0=16t^2+183t+46`

`\Rightarrow16t^2+183t+46=0`

Solving the equation we have

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

or

`t=(-183-190.8743)/(32)=(-373.8743)/(32)=-11.685`

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

`t=0.2461\text{ sec}`