Question

a 1.5lb ball is launched out of a canon at 18 ft/s at an angle of 35 degrees above the horizontal. how far away from the canon does the ball land?

Answer 1

The range for projectile motion

$$R=\frac{v_i^2\sin2\theta}{g}$$ $$=\frac{(18\:\rm ft/s)^2\times \sin 70^{\circ}}{32.17\:\rm ft/s^2}=9.64\:\rm ft.$$