The height of a ball after being dropped from a point 110 feet above the ground is given by \( h=-16 t^{2}+110 \), where \( t \) is the time in seconds since the ball was dropped and \( h \) is in feet. Round all answers to 2 decimal places, where necessary. a. When will the ball be 83 feet above the ground? After beconds. When wil the ball reach the ground? After c. For what values of \( t \) does the problem make sense (from a physical standpoint)? Answer in interval notation.

Let's analyze the problem step by step. Given the height function: \[ h(t) = -16t^2 + 110 \] where: - \( h(t) \) is the height in feet, - \( t \) is the time in seconds after the ball is dropped. ### a. When will the ball be **83 feet** above the ground? Set \( h(t) = 83 \) and solve for \( t \): \[ -16t^2 + 110 = 83 \] \[ -16t^2 = 83 - 110 \] \[ -16t^2 = -27 \] \[ t^2 = \frac{27}{16} \] \[ t = \sqrt{\frac{27}{16}} = \frac{3\sqrt{3}}{4} \approx 1.30 \text{ seconds} \] **Answer:** The ball is 83 feet above the ground after **1.30 seconds**. --- ### b. When will the ball **reach the ground**? The ball reaches the ground when \( h(t) = 0 \). \[ -16t^2 + 110 = 0 \] \[ -16t^2 = -110 \] \[ t^2 = \frac{110}{16} = 6.875 \] \[ t = \sqrt{6.875} \approx 2.62 \text{ seconds} \] **Answer:** The ball reaches the ground after **2.62 seconds**. --- ### c. For what values of \( t \) does the problem make sense (from a physical standpoint)? Physically, time \( t \) cannot be negative, and the motion is only valid from the moment the ball is dropped until it hits the ground. From part **b**, we know the ball hits the ground at \( t \approx 2.62 \) seconds. **Answer:** The values of \( t \) are in the interval **\[0, 2.62\]**.