For a projectile with speed $u$ and angle of projection $\theta$, the horizontal range is $R=\frac{u^2\sin 2\theta}{g}.$

Since both bodies have the same speed and the same range, $\sin 2\theta_1 = \sin 2\theta_2.$ One angle is given as $\theta_1=60^\circ.$ So, $\sin 2\theta_1 = \sin 120^\circ = \sin 60^\circ.$ For projectiles with the same speed and same range, the two projection angles are complementary: $\theta_2 = 90^\circ-\theta_1 = 30^\circ.$

Maximum height of a projectile is $H=\frac{u^2\sin^2\theta}{2g}.$ Given $u=40\ \text{m/s}$ and $g=10\ \text{m/s}^2$, H=\frac{40^2\sin^2\theta}{2\cdot 10}=rac{1600\sin^2\theta}{20}=80\sin^2\theta.

For $\theta=60^\circ$, $H_1=80\sin^2 60^\circ=80\cdot \frac{3}{4}=60\ \text{m}.$

For $\theta=30^\circ$, $H_2=80\sin^2 30^\circ=80\cdot \frac{1}{4}=20\ \text{m}.$

Therefore, the sum of maximum heights is $H_1+H_2=60+20=80\ \text{m}.$