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?

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

To find the acceleration due to gravity that would make the object reach the ground in half the time, we need to set up an equation using the kinematic equation for motion:

s = ut + (1/2)gt^2

where:
s = height of the tower (400 ft)
u = initial velocity (0 ft/sec, since the object is dropped)
g = acceleration due to gravity
t = time taken to reach the ground (5 sec)

For the first scenario, where the object takes 5 seconds to reach the ground, the equation becomes:

400 = 0(5) + (1/2)g(5)^2

Simplifying this equation:

400 = (1/2)g(25)
800 = 25g

Dividing both sides by 25:

g = 800/25
g = 32 ft/sec^2

Therefore, the acceleration due to gravity in the first scenario is 32 ft/sec^2.

Now, for the second scenario where we need to find the acceleration required for the object to hit the ground in half the time, let's call the new time "t2". We know that "t2" is half of the original time, so "t2" = 5/2 = 2.5 seconds.

Using the same formula, the equation becomes:

400 = 0(2.5) + (1/2)g(2.5)^2

Simplifying this equation:

400 = (1/2)g(6.25)
800 = 6.25g

Dividing both sides by 6.25:

g = 800/6.25
g = 128 ft/sec^2

Therefore, the acceleration due to gravity that would make the object reach the ground in half the time (2.5 seconds) is 128 ft/sec^2.