v = Vi - 32 t
0 = 96 - 32 t
t = 3 s
h = Vi t - 16 t^2
h = 96(3) - 16(3)^2
h = 288 - 144 = 144 ft
A body is thrown vertically upward from the ground with an initial velocity of 96 ft/sec. Find the maximum height attain by the body
2 answers
s-maximum height attained by the body in feet
Vo-initial velocity(ft|sec)
t-time(sec)
s=96t-16t^2
16t^2=96t
16t^2/16=96t/16
t^2=6t
t^2-6t=0
t^2-6t+9=0
(t-3)(t-3)=0
t=3 sec.
s=96t-16t^2
s=96(3)-16(3)^2
s=288-144
s=144 feet
Vo-initial velocity(ft|sec)
t-time(sec)
s=96t-16t^2
16t^2=96t
16t^2/16=96t/16
t^2=6t
t^2-6t=0
t^2-6t+9=0
(t-3)(t-3)=0
t=3 sec.
s=96t-16t^2
s=96(3)-16(3)^2
s=288-144
s=144 feet