How long it takes the arrow to come back to the ground if Andrew is an avid archer and he launches an arrow that takes a parabolic path, modeled by the equation $y = - 4.9 t^{2} + 48 t$ $y=-4.9t^2+48t$ ?

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Answer 1 · 2014-11-28T21:24:04

Since $y = 0$ $y=0$ at the ground level, we have

$0 = - 4.9 t^{2} + 48 t = t (- 4.9 t + 48)$ $0=-4.9t^2+48t=t(-4.9t+48)$

$=> {(t=0),(t=48/4.9 approx 9.8):}$

Hence, it takes approximately $9.8$ s.

I hope that this was helpful.

How long it takes the arrow to come back to the ground if Andrew is an avid archer and he launches an arrow that takes a parabolic path, modeled by the equation y=-4.9t^2+48ty=−4.9t2+48ty=-4.9t^2+48t?