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–96
–64
–32
0
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Do not post any more until you have shown your work on your previously posted questions!
The position function is given by s(t) = -16t^2 + 144. To find the derivative, we differentiate the function with respect to time (t):
s'(t) = d/dt (-16t^2 + 144)
To differentiate -16t^2, we use the power rule which states that if we have a function of the form f(t) = ct^n, then the derivative is given by f'(t) = nct^(n-1).
Therefore, differentiating -16t^2 gives us -32t.
The constant term 144 does not depend on time (t), so its derivative is zero.
Taking the derivative of the position function, we get:
s'(t) = -32t + 0
s'(t) = -32t
Now, we need to find the time t when the paper clip hits the ground. We know that the height of the tower is 144 ft, and the initial velocity of the paper clip is 16 ft/sec.
When the paper clip hits the ground, its position s(t) will be 0. We can set up an equation:
0 = -16t^2 + 144
Rearranging this equation, we get:
16t^2 = 144
Dividing both sides by 16:
t^2 = 9
Taking the square root of both sides:
t = ±3
Since time cannot be negative, we take the positive solution:
t = 3
Now that we have the time when the paper clip hits the ground, we can substitute this value into the velocity function to find the velocity at that time:
s'(3) = -32(3)
s'(3) = -96
Therefore, the velocity of the paper clip when it hits the ground is -96 ft/sec. Thus, the correct answer is –96.