The ranges and heights for two projectiles projected with the same initial velocity at angles 42° and 48° with the horizontal are $$R_1$$, $$R_2$$ and $$H_1$$, $$H_2$$ respectively. Choose the correct option:

We start by recalling the standard results for a projectile launched with an initial speed $$u$$ at an angle $$\theta$$ to the horizontal.

Formula for horizontal range: $$R = \dfrac{u^{2}\sin 2\theta}{g}.$$

Formula for maximum height: $$H = \dfrac{u^{2}\sin^{2}\theta}{2g}.$$

Both projectiles are thrown with the same speed $$u$$, so $$u$$ and $$g$$ are common for the two cases. Let us denote the first angle by $$\theta_{1}=42^{\circ}$$ and the second by $$\theta_{2}=48^{\circ}.$$ Their corresponding ranges and heights are $$R_{1},\,H_{1}$$ and $$R_{2},\,H_{2}.$$

Evaluating the ranges

For the first projectile we have

$$R_{1}= \dfrac{u^{2}\sin 2\theta_{1}}{g} = \dfrac{u^{2}\sin(2\times 42^{\circ})}{g} = \dfrac{u^{2}\sin 84^{\circ}}{g}.$$

For the second projectile we obtain

$$R_{2}= \dfrac{u^{2}\sin 2\theta_{2}}{g} = \dfrac{u^{2}\sin(2\times 48^{\circ})}{g} = \dfrac{u^{2}\sin 96^{\circ}}{g}.$$

Now we use the trigonometric identity $$\sin(180^{\circ}-\alpha)=\sin\alpha.$$ Since $$96^{\circ}=180^{\circ}-84^{\circ},$$ we get $$\sin 96^{\circ}=\sin 84^{\circ}.$$ Substituting this equality above,

$$R_{2}= \dfrac{u^{2}\sin 96^{\circ}}{g}= \dfrac{u^{2}\sin 84^{\circ}}{g}=R_{1}.$$

Thus $$R_{1}=R_{2}.$$

Evaluating the heights

For the first projectile:

$$H_{1}=\dfrac{u^{2}\sin^{2}\theta_{1}}{2g} =\dfrac{u^{2}\sin^{2}42^{\circ}}{2g}.$$

For the second projectile:

$$H_{2}=\dfrac{u^{2}\sin^{2}\theta_{2}}{2g} =\dfrac{u^{2}\sin^{2}48^{\circ}}{2g}.$$

Because $$48^{\circ}>42^{\circ}$$ and $$\sin\theta$$ increases from $$0^{\circ}$$ to $$90^{\circ},$$ we have $$\sin 48^{\circ}>\sin 42^{\circ}.$$ Squaring preserves the inequality, so $$\sin^{2}48^{\circ}>\sin^{2}42^{\circ}.$$ Substituting this into the expressions for $$H_{1}$$ and $$H_{2}$$ gives

$$H_{2}>H_{1}.$$

Combining both comparisons, we find

$$R_{1}=R_{2}\quad\text{and}\quad H_{1}<H_{2}.$$

These results match Option B.

Hence, the correct answer is Option B.