An object is launched on the moon at 16 feet per second (f/s) from a 1152-foot tall tower. The equation for the object's height s at time t seconds after launch is s(t) = -16t^2 + 16t + 1152, where s is in feet. When does the object strike the moon’s surface and why?

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

for the first question it would be the second answer for the second question it would be the third answer and for the third question it would be the first answer

step-by-step explanation:

triangle and cube

step-by-step explanation: