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An arrow is shot with an initial upward velocity of 150 feet per second from a height of 5 feet above the ground. The equation h=−16t2+150t+5 models the height in feet t seconds after the arrow is shot. After the arrow passes its maximum height, it comes down and hits a target that was placed 20 feet above the ground. About how long after the arrow was shot does it hit its intended target?

 Sep 29, 2020
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Calculate h = 20

20 = -16t^2 + 150t + 5      solve for t   using Quadratic Formula

  you will get two values....one is arrow going up....the other (larger) is arrow comong down...

 Sep 29, 2020

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