Two particles are projected in air with speeds $υ_{1} a n d u_{2}$ at angles $θ_{1} a n d θ_{2}$ (both acute and $θ_{1} > θ_{2}$ ) to the horizontal, respectively. If the height reached by the first particle is equal to that of the second , then tick the right choices

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$u_{1} > u_{2}$

time of flight : $T_{1} = T_{2}$

horizontal range : $R_{1} < R_{2}$

$u_{1} < u_{2}$

answer is B, C, D.

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Detailed Solution

$H e i g h t, h = \frac{v_{0}^{2} \sin^{2} θ}{2 g}, i . e ., h \propto \sin^{2} θ$

$∴ \frac{h_{1}}{h_{2}} = \frac{{(u_{1})}^{2} \sin^{2} θ_{1}}{{(u_{2})}^{2} \sin^{2} θ_{2}} = 1$

$S o, u_{1} \sin θ_{1} = u_{2} \sin θ_{2}$

$Since θ_{1} > θ_{2} w e c a n w r i t e u_{1} < u_{2}$

$A l s o t i m e o f f l i g h t, T = \frac{2 u \sin θ}{g}$

So $T_{1} = T_{2}$

$H o r i z o n t a l r a n g e, R = u \cos θ \times T S i n c e T_{1} = T_{2,} w e c a n w r i t e R_{1} \infty u_{1} \cos θ_{1} a n d R_{1} \infty u_{2} \cos θ_{2} A l s o \sin c e θ_{1} > θ_{2} a n d u_{1} < u_{2}, w e c a n w r i t e R_{1} < R_{2} .$