If T is the total time of flight,h is the max | vedclass

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(D) For a projectile launched with initial velocity $u$ at an angle $	heta$ with the horizontal:$1$. The horizontal displacement is $x = (u \cos 	he

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Both $(A)$ and $(B)$

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(D) For a projectile launched with initial velocity $u$ at an angle $	heta$ with the horizontal:$1$. The horizontal displacement is $x = (u \cos 	heta)t$. Since $R = (u \cos 	heta)T$,we have $x/R = t/T$.$2$. The vertical displacement is $y = (u \sin 	heta)t - \frac{1}{2}gt^2$. Since $h = \frac{u^2 \sin^2 	heta}{2g}$ and $T = \frac{2u \sin 	heta}{g}$,we can write $u \sin 	heta = \frac{gT}{2}$.$3$. Substituting $u \sin 	heta$ and $g = \frac{2u \sin 	heta}{T}$ into the equation for $y$:$y = \left( \frac{gT}{2} ight)t - \frac{1}{2}g t^2 = \frac{gT^2}{2} \left( \frac{t}{T} ight) - \frac{gT^2}{2} \left( \frac{t}{T} ight)^2 = \frac{gT^2}{2} \left( \frac{t}{T} ight) \left( 1 - \frac{t}{T} ight)$.$4$. Since $h = \frac{(u \sin 	heta)^2}{2g} = \frac{(gT/2)^2}{2g} = \frac{gT^2}{8}$,we have $\frac{gT^2}{2} = 4h$.$5$. Thus,$y = 4h \left( \frac{t}{T} ight) \left( 1 - \frac{t}{T} ight)$.$6$. Substituting $t/T = x/R$,we get $y = 4h \left( \frac{x}{R} ight) \left( 1 - \frac{x}{R} ight)$.Therefore,both $(A)$ and $(B)$ are correct.

Column $I$	Column $II$
$(A)$ Displacement	$(p)$ $8 \sin 2$
$(B)$ Distance	$(q)$ $4$
$(C)$ Average velocity	$(r)$ $2 \sin 2$
$(D)$ Average acceleration	$(s)$ $4 \sin 2$

If $T$ is the total time of flight,$h$ is the maximum height,and $R$ is the horizontal range,how are the $x$ and $y$ coordinates of projectile motion related to time $t$ and range $R$?

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