Oh my, feet and inches.
Well g= about -32 ft/second^2
so
v = Vi + a t = 160 - 32 t
h = Hi + Vi t + (a/2) t^2 = 70 + 160 t - 16 t^2
That is about it because you did not say what the question is.
A projectile is launched upward with a velocity of 160
160
feet per second from the top of a 70
70
-foot platform.
2 answers
probably want maximum height, and time to reach ground
height(t) = -16t2 + 160t + 70
the t of the vertex = -160/(-32) = 5 seconds
height(5) = -16(25) + 160(5) + 70 = 470 ft
when is -16t2 + 160t + 70 = 0 ?
8t^2 - 80t - 35 = 0
t = (80 ± √(80^2 - 4(8)(-35))/16
= appr 10.42 seconds or a negative time , reject that negative
height(t) = -16t2 + 160t + 70
the t of the vertex = -160/(-32) = 5 seconds
height(5) = -16(25) + 160(5) + 70 = 470 ft
when is -16t2 + 160t + 70 = 0 ?
8t^2 - 80t - 35 = 0
t = (80 ± √(80^2 - 4(8)(-35))/16
= appr 10.42 seconds or a negative time , reject that negative