Jesse tossed a paint brush off her roof. The height of the brush (in meters above the ground) \( t \) seconds after Jesse tossed it is modeled by \( h(t)=-5 t^{2}+5 t+10 \) Jesse wants to know when the brush will hit the ground. 1) Rewrite the function in a different form (factored or vertex) where the answer appears as a number in the equation. \( h(t)=\square \) 2) How many seconds after being thrown does the brush hit the ground? seconds

Let's transform the quadratic function into vertex form by completing the square. 1) We start with \( h(t) = -5t^2 + 5t + 10 \). Factoring out -5 gives us: \[ h(t) = -5(t^2 - t - 2) \] Next, completing the square inside the parentheses: \[ = -5\left((t - \frac{1}{2})^2 - \frac{1}{4} - 2\right) \] \[ = -5\left((t - \frac{1}{2})^2 - \frac{9}{4}\right) \] \[ = -5(t - \frac{1}{2})^2 + \frac{45}{4} \] So the function in vertex form is: \[ h(t) = -5(t - \frac{1}{2})^2 + \frac{45}{4} \] 2) To find when the brush hits the ground, we set \( h(t) = 0 \): \[ -5(t - \frac{1}{2})^2 + \frac{45}{4} = 0 \] Solving for \( t \): \[ (t - \frac{1}{2})^2 = \frac{45}{20} = \frac{9}{4} \] Taking the square root gives us \( t - \frac{1}{2} = ±\frac{3}{2} \), leading us to: 1) \( t = 2 \) seconds or 2) \( t = -1 \) seconds (not valid since time can't be negative). Thus, the brush hits the ground after **2 seconds.**