college algebra

Question

A projectile is launched from ground level with an initial velocity of

v0

feet per second. Neglecting air​ resistance, its height in feet t seconds after launch is given by

s=−16t2+v0t.

Find the​ time(s) that the projectile will​ (a) reach a height of

96

ft and​ (b) return to the ground when

v0=112

feet per second.

Answer 1

s= h= -16t^2 +112t = 96

16t^2-112t +96 = 0

t^2 -7t + 6 = 0

(t-6)(t-1)= 0

t=1, 6 seconds when height = 96 feet

two different times it reaches 96 feet, after 1 second on the way up and at 6 seconds, on the way down

it hits the ground when

s= h =0 = -16t^2 +112t

16t^2 -112t+0

t^2 -7t = 0

t(t-7)= 0

t= 0 or 7 seconds at ground level