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Question Number 54584 by peter frank last updated on 07/Feb/19

	Answered by tanmay.chaudhury50@gmail.com last updated on 07/Feb/19

	$t r y i n g$ $t_{e c h o f r o m b o t t o m} = \frac{2 (200)}{v_{s o u n d i n a i r}} + \frac{2 h}{v_{s o u n d i n w a t e r}}$ $t_{e c h o f r o m s u r f a c e} = \frac{2 (200)}{v_{s o u n d i n a i r}}$ $t_{b} - t_{s} = 0.24 = \frac{2 h}{v_{s o u n d i n w a t e r}}$ $h = \frac{0.24 \times v_{s o u n d i n w a t e r}}{2}$
	Commented by peter frank last updated on 07/Feb/19

	$t h a n k y o u$
	Answered by mr W last updated on 07/Feb/19

	$t_{1} = 2 h / v_{a i r}$ $t_{2} = 2 h / v_{a i r} + 2 d / v_{w a t e r}$ $Δ t = t_{2} - t_{1} = 2 d / v_{w a t e r}$ $\Rightarrow d = \frac{Δ t v_{w a t e r}}{2} = \frac{0.24 \times 1481}{2} = 178 m$
	Commented by mr W last updated on 07/Feb/19

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