A paper clip is dropped from the top of a 144‐ft tower, with an initial velocity of 16 ft/sec. Its position function is s(t) = −16t^2 + 144. What is its velocity in ft/sec when it hits the ground?
–96
–64
–32
0
To find the velocity of the paper clip when it hits the ground, we need to determine the derivative of the position function with respect to time. The derivative of the position function gives us the rate of change of the position, which is the velocity.
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.
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