1. **State the problem:** We want to find the maximum height a ball can reach when thrown straight up by a pitcher who releases the ball from a height of $h_0 = 6 + 1 = 7$ feet with an initial velocity $v_0 = 135$ feet per second. 2. **Formula:** The height function is given by $$h(t) = -16t^2 + v_0 t + h_0$$ where $t$ is time in seconds. 3. **Find the time at maximum height:** The maximum height occurs at the vertex of the parabola. The time $t$ at the vertex is given by $$t = -\frac{b}{2a}$$ where $a = -16$ and $b = v_0 = 135$. 4. Calculate $$t = -\frac{135}{2 \times (-16)} = \frac{135}{32} = 4.21875 \text{ seconds}.$$ 5. **Calculate the maximum height:** Substitute $t = 4.21875$ into the height function: $$h(4.21875) = -16(4.21875)^2 + 135(4.21875) + 7.$$ 6. Compute step-by-step: $$-16(4.21875)^2 = -16 \times 17.799 = -284.784,$$ $$135 \times 4.21875 = 569.531,$$ 7. Sum all terms: $$h(4.21875) = -284.784 + 569.531 + 7 = 291.747 \text{ feet}.$$ 8. **Round to nearest hundredth:** The maximum height is approximately **291.75 feet**. **Final answer:** The maximum height the ball can reach is **291.75 feet**.