The height (feet) of an object moving vertically is given by \( s=-16 t^{2}+176 t+132 \), where \( t \) is in seconds. Find the object's velocity at \( t=5 \), its maximum height and when it occurs, and its velocity when \( s=0 \).

To analyze the motion of the object described by the height function \( s(t) = -16t^2 + 176t + 132 \), we will perform the following steps: 1. **Find the object's velocity at \( t = 5 \)**. 2. **Determine the maximum height and when it occurs**. 3. **Find the velocity when \( s = 0 \)**. ### Step 1: Find the velocity at \( t = 5 \) The velocity \( v(t) \) is the derivative of the height function \( s(t) \): \[ v(t) = \frac{ds}{dt} = \frac{d}{dt}(-16t^2 + 176t + 132) \] Calculating the derivative: \[ v(t) = -32t + 176 \] Now, substituting \( t = 5 \): \[ v(5) = -32(5) + 176 = -160 + 176 = 16 \text{ feet per second} \] ### Step 2: Determine the maximum height and when it occurs The maximum height occurs at the vertex of the parabola described by the quadratic function. The time \( t \) at which the maximum height occurs can be found using the formula: \[ t = -\frac{b}{2a} \] where \( a = -16 \) and \( b = 176 \): \[ t = -\frac{176}{2(-16)} = \frac{176}{32} = 5.5 \text{ seconds} \] Now, we substitute \( t = 5.5 \) back into the height function to find the maximum height: \[ s(5.5) = -16(5.5)^2 + 176(5.5) + 132 \] Calculating \( (5.5)^2 \): \[ (5.5)^2 = 30.25 \] Now substituting: \[ s(5.5) = -16(30.25) + 176(5.5) + 132 \] \[ = -484 + 968 + 132 \] \[ = 616 \text{ feet} \] Thus, the maximum height is \( 616 \) feet, occurring at \( t = 5.5 \) seconds. ### Step 3: Find the velocity when \( s = 0 \) To find when the object hits the ground, we set \( s(t) = 0 \): \[ -16t^2 + 176t + 132 = 0 \] We can use the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \): Here, \( a = -16 \), \( b = 176 \), and \( c = 132 \): Calculating the discriminant: \[ b^2 - 4ac = 176^2 - 4(-16)(132) \] \[ = 30976 + 8448 = 39424 \] Now applying the quadratic formula: \[ t = \frac{-176 \pm \sqrt{39424}}{2(-16)} \] Calculating \( \sqrt{39424} \): \[ \sqrt{39424} = 198.56 \text{ (approximately)} \] Now substituting back into the formula: \[ t = \frac{-176 \pm 198.56}{-32} \] Calculating the two possible values for \( t \): 1. \( t = \frac{-176 + 198.56}{-32} = \frac{22.56}{-32} \approx -0.705 \) (not valid since time cannot be negative) 2. \( t = \frac{-176 - 198.56}{-32} = \frac{-374.56}{-32} \approx 11.7 \text{ seconds} \) Now, we find the velocity at \( t \approx 11.7 \): \[ v(11.7) = -32(11.7) + 176 \] \[ = -374.4 + 176 = -198.4 \text{ feet per second} \] ### Summary of Results 1. The object's velocity at \( t = 5 \) is \( 16 \) feet per second. 2. The maximum height is \( 616 \) feet, occurring at \( t = 5.5 \) seconds. 3. The velocity when \( s = 0 \) (when it hits the ground) is approximately \( -198.4 \) feet per second.