Chapter 13: Vector-Valued Functions and Motion in Space - Section 13.2 - Integrals of Vector Functions; Projectile Motion - Exercises 13.2 - Page 753: 20

$$490 \mathrm{m} / \mathrm{s}$$

Since $$R=\frac{v_{0}^{2}}{g} \sin 2 \alpha$$ and the maximum $R$ occurs when $\alpha=45^{\circ} $ Then \begin{align*} 24.5 \mathrm{km}&=\left(\frac{v_{0}^{2}}{9.8 \mathrm{m} / \mathrm{s}^{2}}\right)\left(\sin 90^{\circ}\right)\\ v_{0}&=\sqrt{(9.8)(24,500) \mathrm{m}^{2} / \mathrm{s}^{2}}=490 \mathrm{m} / \mathrm{s}\end{align*}