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Question Number 211418 by peter frank last updated on 08/Sep/24

	Answered by Spillover last updated on 09/Sep/24

	Answered by Spillover last updated on 10/Sep/24

	$P a r t i c l e A$ $H o r i z o n t a l d i s t a n c e, x_{A} = U_{A} t \cos α$ $V e r t i c a l d i s t a n c e, y_{A} = h + U_{A} t \sin α - \frac{1}{2} {g t}^{2}$ $P a r t i c l e B$ $H o r i z o n t a l d i s t a n c e, x_{B} = 2 h + U_{B} t \cos β$ $V e r t i c a l d i s t a n c e, y_{B} = h + U_{B} t \sin β - \frac{1}{2} {g t}^{2}$ $C o l l i s i o n$ $x_{A} = U_{A} t \cos α = x_{B} = 2 h + U_{B} t \cos β$ $t = \frac{2 h}{U_{A} \cos α - U_{B} \cos β}$ $h + U_{A} t \sin α - \frac{1}{2} {g t}^{2} = h + U_{B} t \sin β - \frac{1}{2} {g t}^{2}$ $h = t (U_{B} \sin B - U_{A} \sin α)$ $h = \frac{2 h (U_{B} \sin B - U_{A} \sin α)}{U_{A} \cos α - U_{B} \cos β} .$ $U_{A} \cos α - U_{B} \cos β = (U_{B} \sin B - U_{A} \sin α)$ $\frac{U_{A}}{U_{B}} = \frac{(co s α + 2 \sin α)}{= (\cos β + 2 \sin β)}$
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