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Question Number 133838 by I want to learn more last updated on 24/Feb/21

Commented by mr W last updated on 25/Feb/21

$i t i s n o t c l e a r w h a t i s m e a n t . i s t h e$ $l i n e m a n 4.6 a w a y f r o m t h e g o a l p o s t s$ $o r f r o m t h e k i c k e r ? i a s s u m e d h e i s$ $4.6 m a w a y f r o m t h e g o a l p o s t s .$
	Answered by mr W last updated on 25/Feb/21

	Commented by I want to learn more last updated on 24/Feb/21

	$I really appreciate your time sir . Thanks .$
	Commented by mr W last updated on 24/Feb/21

	$L = 45.7 m$ $h = 3 m$ $θ = 45 °$ $(a)$ $t = \frac{L}{u \cos θ}$ $h = \frac{L}{u \cos θ} (u \sin θ - \frac{g}{2} \times \frac{L}{u \cos θ})$ $h = L [\tan θ - \frac{g L}{2 u^{2}} (1 + \tan^{2} θ)]$ $\frac{h}{L} = \tan θ - \frac{g L}{2 u^{2}} (1 + \tan^{2} θ)$ $l e t λ = \frac{g L}{2 u^{2}}, η = \frac{h}{L}, ξ = \tan θ$ $η = ξ - λ (1 + ξ^{2})$ $λ = \frac{ξ - η}{1 + ξ^{2}}$ $\frac{d λ}{d ξ} = \frac{1}{1 + ξ^{2}} - \frac{2 ξ^{2}}{{(1 + ξ^{2})}^{2}} = 0$ $\Rightarrow 1 - \frac{2 ξ^{2}}{1 + ξ^{2}} = 0$ $\Rightarrow ξ^{2} = 1 \Rightarrow \tan θ = 1 \Rightarrow θ = 45 °$ $i . e . a t θ = 45 ° w e h a v e λ_{m a x} o r u_{m i n .}$ $η = 1 - λ_{m a x} (1 + 1) \Rightarrow λ_{m a x} = \frac{1 - η}{2}$ $\frac{g L}{2 u_{m i n}^{2}} = \frac{1 - \frac{h}{L}}{2} = \frac{L - h}{2 L}$ $u_{m i n} = \sqrt{\frac{{g L}^{2}}{L - h}} = \sqrt{\frac{10 \times 45 . 7^{2}}{45.7 - 3}} = 22.116 m / s$ $(b)$ $d = 4.6 m$ $h_{1} = L_{1} [\tan θ - \frac{{g L}_{1}}{2 u^{2}} (1 + \tan^{2} θ)]$ $h_{1} = L_{1} [\tan θ - \frac{L_{1}}{L} \times \frac{g L}{2 u^{2}} (1 + \tan^{2} θ)]$ $h_{1} = L_{1} [1 - \frac{L_{1}}{L} \times \frac{L - h}{2 L} \times 2]$ $h_{1} = L_{1} [1 - \frac{L_{1}}{L} \times \frac{L - h}{L}]$ $a t L_{1} = L - d = 45.7 - 4.6 = 41.1 m :$ $h_{1} = 41.1 \times (1 - \frac{41.1 \times 42.7}{45 . 7^{2}}) = 6.56 m > 2.5 m$ $i . e . t h e l i n e m a n {c a n}^{'} t b l o c k t h e b a l l .$ $(c)$ $d = 1 m$ $L_{1} = 44.7 m$ $h_{1} = 44.7 [1 - \frac{44.7 \times 42.7}{45 . 7^{2}}] = 3.85 m > 2.5 m$ $i . e . t h e l i n e m a n {c a n}^{'} t b l o c k t h e b a l l .$
	Commented by Abdoulaye last updated on 24/Feb/21

	$h ow do you do these kinds of figures$ $on this app ?$
	Commented by mr W last updated on 24/Feb/21

	$u s i n g t h e a p p L e k h .$
	Commented by mr W last updated on 25/Feb/21

	$t h e q u e s t i o n c o u l d a l s o m e a n t h i s :$
	Commented by mr W last updated on 25/Feb/21

	Commented by I want to learn more last updated on 25/Feb/21

	$Ohhh, alright sir . Thanks .$
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