SOLUTION: Science and Medicine: the equation h=16r^2+112t gives the height of an arrow shot upward from the ground with an initial velocity of 112f/s where t is the time after the arrow lea

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Question 49728This question is from textbook
: Science and Medicine: the equation h=16r^2+112t gives the height of an arrow shot upward from the ground with an initial velocity of 112f/s where t is the time after the arrow leaves the ground find the time it takes for the arrow to reach a height of 180fThis question is from textbook

Answer by Earlsdon(6103) About Me  (Show Source):
You can put this solution on YOUR website!
I think that you have omitted a negative sign from the first term ot the equation: The general form for the function describing the height (as a funtion of time, t) of an object propelled upwards is: h%28t%29+=+-16t%5E2%2BVot%2BHo where: Vo is the initial upwards velocity and Ho is the initial height.
In your problem, Vo = 112 ft/sec and Ho = 0 (Ground-level), so your equation should be:
h%28t%29+=+-16t%5E2%2B112t Set h(t) = 180 ft. and solve for t.
180+=+-16t%5E2%2B112t Subtract 180 from both sides of the equation.
-16t%5E2%2B112t-180+=+0 Simplify by factoring out a -4.
-4%284t%5E2-28t%2B45%29+=+0
4t%5E2-28t%2B45+=+0 Solve by factoring.
%282t-5%29%282t-9%29+=+0 Apply the zero product principle.
2t-5+=+0 and/or 2t-9+=+0
If 2t-5+=+0 then 2t+=+5 and t+=+2.5
If 2t-9+=+0 then 2t+=+9 and t+=+4.5
The arrow reaches a height of 180 ft in 2.5 seconds on its way up and it passes the 180 foot-level on its way down in 4.5 seconds.