An object is dropped from the top of a cliff 640 meters high. Its height above the ground \( t \) seconds after it is dropped is \( 640-4.9 \mathrm{t}^{2} \). Determine its speed 8 second affer it is dropped. The speed of the object 8 seconds after it is dropped is \( \square \mathrm{m} / \mathrm{sec} \) (Type an integer or a decimal. Do not round.)

To find the speed of the object after 8 seconds, we first need to understand the relationship between height and speed. The height \( h \) of the object at time \( t \) seconds is given by the equation \( h(t) = 640 - 4.9t^2 \). To find the speed, we need to calculate the derivative of the height function to get the velocity function. The height function \( h(t) \) can be differentiated to find the velocity \( v(t) \): \[ v(t) = \frac{dh}{dt} = \frac{d}{dt}(640 - 4.9t^2) = -9.8t \] Now we can find the speed at \( t = 8 \) seconds: \[ v(8) = -9.8 \times 8 = -78.4 \text{ m/sec} \] Since speed is the magnitude of velocity, we take the absolute value: \[ \text{Speed} = 78.4 \text{ m/sec} \] Thus, the speed of the object 8 seconds after it is dropped is \( 78.4 \text{ m/sec} \).