it is at h(0.5) feet
fix your typo and evaluate that.
h(t)=-16^(2)+50 where is the ball after 0.5 seconds
3 answers
Equation incomplete.
I guess you are dropping a ball using old feet and pound units from 50 feet high.
In those units g is about 32 ft/second^2
in general for constant acceleration a
v = Vinitial + a t
x = Xinitial + Vinitial t + (1/2) a t^2
here
a = -g = -32
Vi = 0
Xi = 50 = h
so
h = 50 + 0 + (1/2)(-32) t^2
if t = 1/2
t^2 = 1/4
h = 50 - 16/4 = 50 - 4 = 46 feet
In those units g is about 32 ft/second^2
in general for constant acceleration a
v = Vinitial + a t
x = Xinitial + Vinitial t + (1/2) a t^2
here
a = -g = -32
Vi = 0
Xi = 50 = h
so
h = 50 + 0 + (1/2)(-32) t^2
if t = 1/2
t^2 = 1/4
h = 50 - 16/4 = 50 - 4 = 46 feet
1 answer