An object falls off a 400-foot tower. With g=32ft/sec^2, as at the Earth’s surface, the object would hit the ground after 5 seconds. What would the acceleration due to gravity have to be to make it reach the ground in half the time?

Answer 1

should have been g = -32
then
v = -32t + c, but when t = 0 , v = 0 , so c = 0

we have v(t) =-32t
then
a(t) = -16t^2 + k, when t = 0 , s = 400 , so k = 400

so our distance equation is
s(t) = -16t^2 + 400
when it hits the ground, s = 0
0 = -16t^2 + 400
16t^2= 400
t^2 = 25
t = 5
which was given, and is correct

if we didn't know the acceleration, let it be a
our distance equation would have been
s(t) = (a/2)t^2 + 400
at half the time, t = 2.5

0 = (a/2)(6.25) + 400
a/2 = -400/6.25
a = -800/6.25 = -128 ft/sec^2