The height (in feet) of a rocket from ground level is given by the function f(t) = -16t^2 + 160t. What is the instantaneous velocity of the rocket 3 seconds after it is launched?

v(t) = df/dt = -32t+160

To find the instantaneous velocity of the rocket 3 seconds after it is launched, we need to find the derivative of the given function, f(t), with respect to time, t. The derivative of f(t) represents the rate of change of the rocket's height with respect to time, which is the velocity.

To find the derivative, we can differentiate each term of the function separately using the power rule and the product rule of differentiation.

Given: f(t) = -16t^2 + 160t

Differentiating the first term:
d/dt (-16t^2) = -16 * 2t = -32t

Differentiating the second term:
d/dt (160t) = 160

Combining the derivatives, we get:
f'(t) = -32t + 160

This function represents the instantaneous velocity of the rocket at any given time, t. Now, we can substitute the value t = 3 seconds into the velocity function to find the instantaneous velocity 3 seconds after the rocket is launched.

Substituting t = 3 into f'(t):
f'(3) = -32(3) + 160
f'(3) = -96 + 160
f'(3) = 64

Therefore, the instantaneous velocity of the rocket 3 seconds after it is launched is 64 feet per second.