Question

From a clifftop over the ocean 592.9 m above sea level, an object was shot straight up into the air with an initial vertical speed of 94.08 ms. On its way down it missed the cliff and fell into the ocean, where it floats on the surface. Its height (above sea level) as time passes can be modeled by the quadratic function f, where f(t)=−4.9t2+94.08t+592.9. Here t represents the number of seconds since the object’s release, and f(t) represents the object’s height (above sea level) in meters. Find the function’s domain and range in this context.

The function’s domain in this context is

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Equation Editor . The function’s range in this context is

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Answer 1

From a clifftop over the ocean 592.9 m above sea level, an object was shot straight up into the air with an initial vertical speed of 94.08 ms. On its way down it missed the cliff and fell into the ocean, where it floats on the surface. Its height (above sea level) as time passes can be modeled by the quadratic function f, where f(t)=-49t2+94.08t+5929. Here t represents the number of seconds since the object's release, and fit) represents the object's height (above sea level) in meters. Find the function's domain and range in this context. (9.6, 1044.484) 1000 Iif we ne 20. (0, 592.9) (24.2, 0) 20 30 Neglest legative value Го, 24, 27 ラ 2o 久a Verkx- ( 6, loyy.yy