1. For a projectile projected with speed $u$ at angle $\theta$:

   $R = \frac{u^2 \sin 2\theta}{g}$ $H = \frac{u^2 \sin^2 \theta}{2g}$

2. Compare the ranges for $\theta_1 = 42^\circ$ and $\theta_2 = 48^\circ$.

   $R_1 = \frac{u^2 \sin(2\cdot 42^\circ)}{g} = \frac{u^2 \sin 84^\circ}{g}$ $R_2 = \frac{u^2 \sin(2\cdot 48^\circ)}{g} = \frac{u^2 \sin 96^\circ}{g}$

   Now,

   $\sin 96^\circ = \sin(180^\circ - 96^\circ) = \sin 84^\circ$

   Hence,

   $R_1 = R_2$

   This also follows from the fact that $42^\circ + 48^\circ = 90^\circ$, so complementary angles give equal range for the same speed.

3. Compare the maximum heights.

   $H_1 = \frac{u^2 \sin^2 42^\circ}{2g}$ $H_2 = \frac{u^2 \sin^2 48^\circ}{2g}$

   Since $48^\circ > 42^\circ$ and $\sin \theta$ increases in this interval,

   $\sin 48^\circ > \sin 42^\circ$

   Therefore,

   $\sin^2 48^\circ > \sin^2 42^\circ$

   so,

   $H_2 > H_1$

   Hence,

   $H_1 < H_2$

4. Final conclusion:

   $R_1 = R_2 \quad \text{and} \quad H_1 < H_2$

Therefore, the correct option is B.