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
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).
1 answer
1 answer