Asked by Greg
An object is projected directly upward from the ground. Its distance in feet from the ground in t seconds is s equals s=112t−16t^2
The object will be 160 feet from the ground after () second(s) and after
() second(s).
The object strikes the ground after () second(s).
The object will be 160 feet from the ground after () second(s) and after
() second(s).
The object strikes the ground after () second(s).
Answers
Answered by
Damon
16 t^2 -112 t +160 = 0
t^2 - 7 t + 10 = 0
(t-5)(t-2) = 0
so at 2 seconds on the way up and at 5 seconds on the way down
0 = 16t (7 - t)
crash at t = 7
t^2 - 7 t + 10 = 0
(t-5)(t-2) = 0
so at 2 seconds on the way up and at 5 seconds on the way down
0 = 16t (7 - t)
crash at t = 7
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