Asked by Kimberly
A projectile was launched from ground level with an initial velocity of the v0 neglecting air resistance is high in feet two seconds after launch is given by S= -16T^2 + v0t find the times that the projectile will reach a high of 128 feet and return to the ground when the zero equals 96 feet per second
Answers
Answered by
oobleck
what does "the zero equals 96 feet per second" mean?
I think you meant to say v0 = 96 ft/s
s = -16t^2 + 96t
so you need to solve
-16t^2 + 96t = 128
t^2 - 6t + 8 = 0
(t-2)(t-4) = 0
t=2 (on the way up)
t=4 (on the way back down)
and
-16t^2 + 96t = 0
-16t(t-6) = 0
t=0 (at the start)
t=6 (upon impact)
I think you meant to say v0 = 96 ft/s
s = -16t^2 + 96t
so you need to solve
-16t^2 + 96t = 128
t^2 - 6t + 8 = 0
(t-2)(t-4) = 0
t=2 (on the way up)
t=4 (on the way back down)
and
-16t^2 + 96t = 0
-16t(t-6) = 0
t=0 (at the start)
t=6 (upon impact)
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