1. **State the problem:** We have the height function of a container dropped from 25 feet given by $$h = -16t^2 + 25$$ where $h$ is height in feet and $t$ is time in seconds. 2. **Part A: Find the height after 0.5 seconds.** Use the formula by substituting $t=0.5$: $$h = -16(0.5)^2 + 25$$ Calculate the square: $$h = -16(0.25) + 25$$ Multiply: $$h = -4 + 25$$ Add: $$h = 21$$ So, the height after 0.5 seconds is 21 feet. 3. **Part B: Find the time when the container reaches the ground.** The container reaches the ground when height $h=0$: $$0 = -16t^2 + 25$$ Rearrange: $$16t^2 = 25$$ Divide both sides by 16: $$t^2 = \frac{25}{16}$$ Express with cancellation: $$t^2 = \frac{\cancel{25}}{\cancel{16}}$$ (no common factors to cancel here, so just write as is) Take the square root of both sides: $$t = \pm \sqrt{\frac{25}{16}} = \pm \frac{5}{4} = \pm 1.25$$ Since time cannot be negative, we take the positive value: $$t = 1.25$$ seconds. **Final answers:** - Part A: Height after 0.5 seconds is **21 feet**. - Part B: Time to reach the ground is **1.25 seconds**.