Question-55191

Question Number 55191 by Tawa1 last updated on 19/Feb/19

Answered by tanmay.chaudhury50@gmail.com last updated on 19/Feb/19

18) a t g r e a t e s t h e u g h t v e r t i c a l c o m p o n e n t o f

v e l o c i t y i s z e r o

0^{2} = {(u s i n θ)}^{2} - 2 {g h}_{1}

0^{2} = {u s i n (90 - θ)}^{2} - 2 {g h}_{2}

\frac{2 {g h}_{1}}{2 {g h}_{2}} = \frac{u^{2} {s i n}^{2} θ}{u^{2} {c o s}^{2} θ}

t a n θ = \sqrt{\frac{h_{1}}{h_{2}}}

Commented by Tawa1 last updated on 19/Feb/19

God bless you sir

Answered by tanmay.chaudhury50@gmail.com last updated on 19/Feb/19

19) R a n g e = (u c o s θ) \times (t i m e o f f l i g h t)

t i m e o f f l i g h t = T = 2 t_{0}

0 = u s i n θ - {g t}_{0}

t_{0} = \frac{u s i n θ}{g}

R = (u c o s θ) (\frac{2 u s i n θ}{g}) = \frac{u^{2} s i n 2 θ}{g}

R_{m a x} w h e n s i n 2 θ = 1 = s i n \frac{π}{2}

θ = \frac{π}{4}

{s i n}^{- 1} (\frac{\sqrt{2}}{2}) \to {s i n}^{- 1} (\frac{1}{\sqrt{2}}) \to \frac{π}{4}

Commented by Tawa1 last updated on 19/Feb/19

God bless you sir

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