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Question Number 178582 by Best1 last updated on 19/Oct/22

s h o w t h a t R a n g e o f t h e f f p r o j e c t i o n

o b t a i n e d b y a l g e b r i c e x p r e s s i o n

R = \frac{(u c o s θ) (u s i n θ) + \sqrt{{(u s i n θ)}^{2} + 2 g h}}{g} h e l p m e p l e a s e

Commented by Best1 last updated on 18/Oct/22

p l e a s e h e l p m e

Commented by Ar Brandon last updated on 18/Oct/22

dimension of (\sin θ^{2}) \neq dimension of 2 g h = M^{2} T^{- 2}

Check question .

Commented by Best1 last updated on 19/Oct/22

i c o r r e c t e d i t n o w h e l p m e

Answered by Ar Brandon last updated on 18/Oct/22

For horizontal displacement

u_{x} = u \cos θ \Rightarrow x = u_{x} t

\Rightarrow R = (u \cos θ) t \Rightarrow t = \frac{R}{u \cos θ} \dots (i)

For vertical displacement

y = u_{y} t + \frac{1}{2} {a t}^{2}

\Rightarrow h = (u \sin θ) t - \frac{1}{2} {g t}^{2} \Rightarrow {g t}^{2} - 2 (u \sin θ) t + 2 h = 0

\Rightarrow t = \frac{2 u \sin θ \pm \sqrt{4 u^{2} \sin^{2} θ - 8 g h}}{2 g}

\Rightarrow t = \frac{u \sin θ + \sqrt{u^{2} \sin^{2} θ - 2 g h}}{g} \dots (i i)

(i) = (i i)

\Rightarrow \frac{R}{u \cos θ} = \frac{u \sin θ + \sqrt{u^{2} \sin^{2} θ - 2 g h}}{g}

\Rightarrow R = \frac{(u \sin θ) (u \cos θ) + u \cos θ \sqrt{u^{2} \sin^{2} θ - 2 g h}}{g}

Commented by Best1 last updated on 18/Oct/22

i s t h i s t h e s a m e a s \frac{(u c o s θ) (u s i n θ) + \sqrt{s i n θ^{2} - 2 g h}}{g} ? ? ?

Commented by Ar Brandon last updated on 18/Oct/22

I think this {isn}^{'} t correct as the dimensions

{aren}^{'} t the same .