An object is thrown upward at a speed of 119 feet per second by a machine from a height of 15 feet off the ground. The height of the object after t seconds can be found using the equation h(t)=-16t^2+119t+15 Give all numerical answers to 2 decimal places. When will the height be 141 feet? At square units: square When will the object reach the ground? square units: square When will the object reach its maximum height? square units: square What is its maximum height? square units: square

When will the height be 141 feet? At 0.56 units: 14.06 ### When will the object reach the ground? 7.58 units: ### When will the object reach its maximum height? 3.72 units: ### What is its maximum height? 236.19 units: Explanation 1. Solve for when height is 141 feet Set h(t) = 141: -16t^2 + 119t + 15 = 141. Simplify to -16t^2 + 119t - 126 = 0. Use the quadratic formula t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} with a = -16, b = 119, c = -126. 2. Calculate discriminant and roots Discriminant \Delta = 119^2 - 4(-16)(-126) = 14161 - 8064 = 6097. Roots are t = \frac{-119 \pm \sqrt{6097}}{-32}. 3. Evaluate roots t_1 = \frac{-119 + \sqrt{6097}}{-32} \approx 0.56, t_2 = \frac{-119 - \sqrt{6097}}{-32} \approx 14.06. 4. Solve for when object reaches the ground Set h(t) = 0: -16t^2 + 119t + 15 = 0. Use the quadratic formula again with a = -16, b = 119, c = 15. 5. Calculate discriminant and roots for ground Discriminant \Delta = 119^2 - 4(-16)(15) = 14161 + 960 = 15121. Roots are t = \frac{-119 \pm \sqrt{15121}}{-32}. 6. Evaluate roots for ground t_1 = \frac{-119 + \sqrt{15121}}{-32} \approx -0.13, t_2 = \frac{-119 - \sqrt{15121}}{-32} \approx 7.58. Only positive root is valid. 7. Find time of maximum height Maximum height occurs at vertex t = \frac{-b}{2a} = \frac{-119}{2(-16)} = 3.72. 8. Calculate maximum height Substitute t = 3.72 into h(t): h(3.72) = -16(3.72)^2 + 119(3.72) + 15 \approx 236.19.