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)?

Simplify the expression by following steps: - step0: Solution: \(-32t\) The velocity function is given by: \[ v(t) = -32t \] Now, we can find the velocity at \( t = 1 \) second: \[ v(1) = -32(1) = -32 \text{ feet per second} \] ### (b) When will it hit the ground? To find when the object hits the ground, we need to solve for \( t \) when \( s(t) = 0 \): \[ -16t^2 + 174 = 0 \] Let's solve this equation. Solve the quadratic equation by following steps: - step0: Solve using square roots: \(-16t^{2}+174=0\) - step1: Add or subtract both sides: \(-16t^{2}=-174\) - step2: Divide both sides: \(\frac{-16t^{2}}{-16}=\frac{-174}{-16}\) - step3: Divide the numbers: \(t^{2}=\frac{87}{8}\) - step4: Simplify the expression: \(t=\pm \sqrt{\frac{87}{8}}\) - step5: Simplify the expression: \(t=\pm \frac{\sqrt{174}}{4}\) - step6: Separate into possible cases: \(\begin{align}&t=\frac{\sqrt{174}}{4}\\&t=-\frac{\sqrt{174}}{4}\end{align}\) - step7: Rewrite: \(t_{1}=-\frac{\sqrt{174}}{4},t_{2}=\frac{\sqrt{174}}{4}\) The solutions to the equation \( -16t^2 + 174 = 0 \) are: \[ t_1 = -\frac{\sqrt{174}}{4}, \quad t_2 = \frac{\sqrt{174}}{4} \] Since time cannot be negative, we will only consider the positive solution: \[ t = \frac{\sqrt{174}}{4} \] Now, let's calculate the numerical value of \( t \). Calculate the value by following steps: - step0: Calculate: \(\frac{\sqrt{174}}{4}\) The time when the object hits the ground is approximately: \[ t \approx 3.30 \text{ seconds} \] ### (c) What is its velocity upon impact (at the instant when it hits the ground)? To find the velocity upon impact, we will substitute \( t = \frac{\sqrt{174}}{4} \) into the velocity function \( v(t) = -32t \): \[ v\left(\frac{\sqrt{174}}{4}\right) = -32\left(\frac{\sqrt{174}}{4}\right) \] Let's calculate this value. Calculate the value by following steps: - step0: Calculate: \(-32\left(\frac{\sqrt{174}}{4}\right)\) - step1: Remove the parentheses: \(-32\times \frac{\sqrt{174}}{4}\) - step2: Reduce the numbers: \(-8\sqrt{174}\) The velocity upon impact is approximately: \[ v \approx -105.53 \text{ feet per second} \] ### Summary of Results: (a) The velocity 1 second after being dropped is \(-32\) feet per second. (b) The object will hit the ground at approximately \(3.30\) seconds. (c) The velocity upon impact is approximately \(-105.53\) feet per second.