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Question Number 217575 by ajfour last updated on 16/Mar/25

	Answered by mahdipoor last updated on 16/Mar/25
	$u_i , θ_i ⇒ d_i =((2u_i ^2 sin(θ_i )cos(θ_i ))/g) u_(i+1) ^→ = u_i (cos(θ_i )i+esin(θ_i )j) = u_(i+1) , θ_(i+1) { ((u_(i+1) ^2 =u_i ^2 (cos^2 (θ_i )+e^2 sin^2 (θ_i )))),((sin(θ_(i+1) )cos(θ_(i+1) )=((esin(θ_i )cos(θ_i ))/(cos^2 (θ_i )+e^2 sin^2 (θ_i ))))) :} d_(i+1) = ed_i Σd=(1+e+e^2 +...)d_0 =((1−e^(oo) )/(1−e))d_0 ⇒ e<1 ⇒ (1/(1−e))×((u_0 ^2 sin(2θ_0 ))/g)$
	$u_{i}, θ_{i} \Rightarrow$ $d_{i} = \frac{2 u_{i}^{2} s i n (θ_{i}) c o s (θ_{i})}{g}$ ${\vec{u}}_{i + 1} = u_{i} (c o s (θ_{i}) i + e s i n (θ_{i}) j) = u_{i + 1}, θ_{i + 1}$ ${\begin{cases} u_{i + 1}^{2} = u_{i}^{2} ({c o s}^{2} (θ_{i}) + e^{2} {s i n}^{2} (θ_{i})) \\ s i n (θ_{i + 1}) c o s (θ_{i + 1}) = \frac{e s i n (θ_{i}) c o s (θ_{i})}{{c o s}^{2} (θ_{i}) + e^{2} {s i n}^{2} (θ_{i})} \end{cases}$ $d_{i + 1} = {e d}_{i}$ $Σ d = (1 + e + e^{2} + . . .) d_{0} = \frac{1 - e^{o o}}{1 - e} d_{0} \Rightarrow e < 1 \Rightarrow$ $\frac{1}{1 - e} \times \frac{u_{0}^{2} s i n (2 θ_{0})}{g}$
	Commented by ajfour last updated on 16/Mar/25

	$e x a m p l e$ $u = 8 m / s$ $g = 10 m / s^{2}$ $θ_{0} = 45 °$ $e = \frac{1}{2}$ $d_{0} = \frac{64}{10} = 6.4 m$ $Σ d = \frac{2 \times 64}{10} = 12.8 m$ $T h a n k y o u . i l l r e v i e w a g a i n .$
	Answered by mr W last updated on 16/Mar/25

	Commented by mr W last updated on 17/Mar/25

	$0 < e < 1$ $u_{n - 1} \cos θ_{n - 1} = u_{n} \cos θ_{n} . . . (i)$ ${e u}_{n - 1} \sin θ_{n - 1} = u_{n} \sin θ_{n} . . . (i i)$ $\frac{u_{n}}{u_{n - 1}} = \frac{\cos θ_{n - 1}}{\cos θ_{n}}$ $\frac{\tan θ_{n}}{\tan θ_{n - 1}} = e$ $l_{n} = \frac{2 u_{n}^{2} \sin θ_{n} \cos θ_{n}}{g}$ $l_{n - 1} = \frac{2 u_{n - 1}^{2} \sin θ_{n - 1} \cos θ_{n - 1}}{g}$ $\frac{l_{n}}{l_{n - 1}} = {(\frac{u_{n}}{u_{n - 1}})}^{2} \frac{\sin θ_{n} \cos θ_{n}}{\sin θ_{n - 1} \cos θ_{n - 1}}$ $= \frac{\cos^{2} θ_{n - 1} \sin θ_{n} \cos θ_{n}}{\cos^{2} θ_{n} \sin θ_{n - 1} \cos θ_{n - 1}}$ $= \frac{\tan θ_{n}}{\tan θ_{n - 1}} = e$ $l_{n} = {e l}_{n - 1} = e^{2} l_{n - 2} = . . . = e^{n} l_{0}$ $w i t h l_{0} = \frac{u_{0}^{2} \sin 2 θ_{0}}{g}$ $L = \underset{n = 0}{\sum^{\infty}} l_{n} = l_{0} \underset{n = 0}{\sum^{\infty}} e^{n} = \frac{l_{0}}{1 - e} = \frac{u_{0}^{2} \sin 2 θ_{0}}{g (1 - e)}$
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