Given below are two statements. One is labell | vedclass

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(A) For the same range,the angles of projection must satisfy $	heta_{1} + 	heta_{2} = 90^{\circ}$,which means $	heta_{2} = 90^{\circ} - 	heta_{1}$

Vedclass · Accepted Answer

Both $A$ and $R$ are true and $R$ is the correct explanation of $A$.

Vedclass · Answer

(A) For the same range,the angles of projection must satisfy $	heta_{1} + 	heta_{2} = 90^{\circ}$,which means $	heta_{2} = 90^{\circ} - 	heta_{1}$.The maximum heights are given by $h_{1} = \frac{u^{2} \sin^{2} 	heta_{1}}{2g}$ and $h_{2} = \frac{u^{2} \sin^{2} 	heta_{2}}{2g} = \frac{u^{2} \cos^{2} 	heta_{1}}{2g}$.Multiplying these heights,we get $h_{1} h_{2} = \left(\frac{u^{2} \sin^{2} 	heta_{1}}{2g}ight) \cdot \left(\frac{u^{2} \cos^{2} 	heta_{1}}{2g}ight)$.This can be rewritten as $h_{1} h_{2} = \frac{u^{4} \sin^{2} 	heta_{1} \cos^{2} 	heta_{1}}{4g^{2}} = \left(\frac{u^{2} \cdot 2 \sin 	heta_{1} \cos 	heta_{1}}{4g}ight)^{2} = \left(\frac{u^{2} \sin(2	heta_{1})}{4g}ight)^{2}$.Since the range $R = \frac{u^{2} \sin(2	heta_{1})}{g}$,we have $h_{1} h_{2} = \left(\frac{R}{4}ight)^{2} = \frac{R^{2}}{16}$.Therefore,$R^{2} = 16 h_{1} h_{2}$,which implies $R = 4 \sqrt{h_{1} h_{2}}$.Both Assertion $A$ and Reason $R$ are true,and $R$ provides the correct mathematical derivation for $A$.

$(A)$ Range	$(i)$ $\frac{P}{Q}$
$(B)$ Maximum height	$(ii)$ $P$
$(C)$ Time of flight	$(iii)$ $\frac{P^2}{4 Q}$
$(D)$ Tangent of projection	$(iv)$ $\left(\sqrt{\frac{2}{g Q}}\right) P$

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