1. **Problem:** Find how high a ball goes, and do an example with a drawing. 2. **Formula used:** For vertical motion, the height is often modeled by $$h(t)=-16t^2+v_0t+h_0$$ where $h(t)$ is height, $v_0$ is the initial upward velocity, and $h_0$ is the starting height. 3. **Important rule:** The ball reaches its highest point at the vertex of the parabola. For $h(t)=at^2+bt+c$, the time of the maximum is $$t=\frac{-b}{2a}$$ Because $a<0$, the parabola opens downward, so the vertex is the top. 4. **Example:** Let the ball be thrown upward from the ground with initial velocity $64$ feet per second. Then $$h(t)=-16t^2+64t$$ Here, $a=-16$ and $b=64$. 5. **Find the time of the highest point:** $$t=\frac{-b}{2a}=\frac{-64}{2\cdot(-16)}=\frac{-64}{-32}=2$$ Using cancellation, $$\frac{\cancel{-64}}{\cancel{-32}}=2$$ So the ball reaches its highest point at $t=2$ seconds. 6. **Find the maximum height:** Substitute $t=2$ into the height formula. $$h(2)=-16(2)^2+64(2)$$ $$h(2)=-16\cdot 4+128$$ $$h(2)=-64+128=64$$ So the ball goes up to a maximum height of $64$ feet. 7. **Final answer:** The ball’s highest point is $64$ feet. 8. **Drawing idea:** The graph is a downward-opening parabola that starts at $0$, rises to the top at $(2,64)$, then comes back down.