Path of projectile is given by the equation Y | vedclass

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Question

(A) The equation of the trajectory is $Y = P x - Q x^2$.Comparing this with the standard equation $Y = x 	an 	heta - \frac{g x^2}{2 u^2 \cos^2 	het

Vedclass · Accepted Answer

$(A)-(i)$,$(B)-(iii)$,$(C)-(iv)$,$(D)-(ii)$

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(A) The equation of the trajectory is $Y = P x - Q x^2$.Comparing this with the standard equation $Y = x 	an 	heta - \frac{g x^2}{2 u^2 \cos^2 	heta}$, we get $	an 	heta = P$ and $Q = \frac{g}{2 u^2 \cos^2 	heta}$.$(A)$ Range $(R)$: At $Y = 0$, $x = R$. So, $0 = P R - Q R^2 \Rightarrow R = \frac{P}{Q}$. Thus, $(A)-(i)$.$(B)$ Maximum height $(H)$: $H = \frac{R 	an 	heta}{4} = \frac{(P/Q) \cdot P}{4} = \frac{P^2}{4 Q}$. Thus, $(B)-(iii)$.$(C)$ Time of flight $(T)$: $T = \frac{2 u \sin 	heta}{g}$. Since $	an 	heta = P$ and $Q = \frac{g}{2 u^2 \cos^2 	heta}$, we have $u \cos 	heta = \sqrt{\frac{g}{2 Q}}$. Then $T = \frac{2 u \sin 	heta}{g} = \frac{2 (u \cos 	heta) 	an 	heta}{g} = \frac{2 \sqrt{\frac{g}{2 Q}} \cdot P}{g} = \left(\sqrt{\frac{2}{g Q}}ight) P$. Thus, $(C)-(iv)$.$(D)$ Tangent of projection: $	an 	heta = P$. Thus, $(D)-(ii)$.

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