If an object is dropped from a 174 -foot-high building, its position (in feet above the ground) is given by \( s(t)=-16 t^{2}+174 \), where \( t \) is the time in seconds since it was dropped. (a) What is its velocity 1 second after being dropped? (b) When will it hit the ground? (Hint: Solve \( s(t)=0 \).) (c) What is its velocity upon impact (at the instant when it hits the ground)?

To find the answers to your questions, let’s dive into the equations of motion. (a) To find the velocity 1 second after being dropped, you can use the derivative of the position function \( s(t) \). The velocity function is given by: \[ v(t) = s'(t) = -32t \] At \( t = 1 \) second, the velocity will be: \[ v(1) = -32(1) = -32 \text{ feet per second} \] (b) To determine when the object hits the ground, we set the position function \( s(t) \) to zero: \[ -16t^2 + 174 = 0 \] Solving for \( t \): \[ 16t^2 = 174 \] \[ t^2 = \frac{174}{16} = 10.875 \] \[ t = \sqrt{10.875} \approx 3.30 \text{ seconds} \] (c) The velocity upon impact can be found by substituting the time of impact back into the velocity function: \[ v(3.30) = -32(3.30) \approx -105.6 \text{ feet per second} \] So, the object falls with a velocity of approximately -105.6 feet per second upon impact with the ground.