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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: 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 180f
This question is from textbook
Answer by Earlsdon(6294) (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: 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:
Set h(t) = 180 ft. and solve for t.
Subtract 180 from both sides of the equation.
Simplify by factoring out a -4.
Solve by factoring.
Apply the zero product principle.
and/or
If then and
If then and
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.
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