An object is dropped from a tower, 156 ft abo a. What is the object's velocity, speed, and ac b. About how long does it take the object to hi c. What is the object's velocity at the moment The object's velocity at time t is

Let’s choose upward as the positive direction. Then: 1. The position of the object at time \( t \) is given by \[ y(t) = 156 - 16t^2, \] where the term \( 16t^2 \) comes from \(\frac{1}{2}(32)t^2\), and \(32\) ft/s\(^2\) is the acceleration due to gravity. 2. The velocity is the derivative of the position: \[ v(t) = \frac{dy}{dt} = -32t. \] Here, the negative sign indicates the velocity is directed downward. The speed is the magnitude of the velocity: \[ \text{Speed} = |v(t)| = 32t. \] The acceleration, being the derivative of the velocity, is constant: \[ a(t) = \frac{dv}{dt} = -32 \text{ ft/s}^2. \] 3. To find when the object hits the ground, set \( y(t)=0 \): \[ 156 - 16t^2 = 0. \] Solving for \( t \): \[ 16t^2 = 156 \quad \Longrightarrow \quad t^2 = \frac{156}{16} = 9.75, \] \[ t = \sqrt{9.75} \approx 3.12 \text{ s}. \] 4. The velocity at impact is found by evaluating \( v(t) \) at \( t \approx 3.12 \) s: \[ v(3.12) = -32(3.12) \approx -99.84 \text{ ft/s}. \] The negative sign confirms that the velocity is downward. Thus, the answers are: a. - Velocity: \( v(t) = -32t \) ft/s, - Speed: \( |v(t)| = 32t \) ft/s, - Acceleration: \( a(t) = -32 \) ft/s\(^2\). b. The object takes approximately \( 3.12 \) seconds to hit the ground. c. The velocity at the moment of impact is approximately \( -100 \) ft/s (downward).