Answer:
0.5seconds later
Step-by-step explanation:
Given the height modeled by the equation:
s(t) = -16t^2 + 16t + 1152
The velocity of the object at the moon surface is zero
ds(t)/dt = 0
ds(t)/dt = -32t + 16
0 = -32t + 16
32 t = 16
t = 16/32
t = 1/2
t = 0.5seconds
Hence the object strikes the moon surface 0.5seconds later