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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?

Sagot :

abidemiokin

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

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