Chapter 13: Vector-Valued Functions and Motion in Space - Section 13.2 - Integrals of Vector Functions; Projectile Motion - Exercises 13.2 - Page 753: 19

$50$ seconds

Since $$x=\left(v_{0} \cos \alpha\right) t $$ Then $$(21 \mathrm{km})\left(\frac{1000 \mathrm{m}}{1 \mathrm{km}}\right)=(840 \mathrm{m} / \mathrm{s})\left(\cos 60^{\circ}\right) t $$ Hence $ t=\frac{21,000 \mathrm{m}}{(840 \mathrm{m} / \mathrm{s})\left(\cos 60^{\circ}\right)}=50$ seconds