Question Number 15832 by Tinkutara last updated on 14/Jun/17
$$\mathrm{A}\:\mathrm{passenger}\:\mathrm{in}\:\mathrm{a}\:\mathrm{train}\:\mathrm{moving}\:\mathrm{at}\:\mathrm{an} \\ $$$$\mathrm{acceleration}\:{a},\:\mathrm{drops}\:\mathrm{a}\:\mathrm{stone}\:\mathrm{from}\:\mathrm{the} \\ $$$$\mathrm{window}.\:\mathrm{A}\:\mathrm{person},\:\mathrm{standing}\:\mathrm{on}\:\mathrm{the} \\ $$$$\mathrm{ground},\:\mathrm{by}\:\mathrm{the}\:\mathrm{sides}\:\mathrm{of}\:\mathrm{the}\:\mathrm{rails}, \\ $$$$\mathrm{observes}\:\mathrm{the}\:\mathrm{ball}\:\mathrm{falling} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{Vertically}\:\mathrm{with}\:\mathrm{an}\:\mathrm{acceleration} \\ $$$$\sqrt{{g}^{\mathrm{2}} \:+\:{a}^{\mathrm{2}} } \\ $$$$\left(\mathrm{2}\right)\:\mathrm{Horizontally}\:\mathrm{with}\:\mathrm{an}\:\mathrm{acceleration} \\ $$$$\sqrt{{g}^{\mathrm{2}} \:+\:{a}^{\mathrm{2}} } \\ $$$$\left(\mathrm{3}\right)\:\mathrm{Along}\:\mathrm{a}\:\mathrm{parabola}\:\mathrm{with}\:\mathrm{an}\:\mathrm{acceleration} \\ $$$$\sqrt{{g}^{\mathrm{2}} \:+\:{a}^{\mathrm{2}} } \\ $$$$\left(\mathrm{4}\right)\:\mathrm{Along}\:\mathrm{a}\:\mathrm{parabola}\:\mathrm{with}\:\mathrm{an}\:\mathrm{acceleration} \\ $$$${g} \\ $$
Answered by mrW1 last updated on 14/Jun/17
$$\mathrm{answer}\:\left(\mathrm{4}\right) \\ $$$$\mathrm{once}\:\mathrm{the}\:\mathrm{stone}\:\mathrm{is}\:\mathrm{droped},\:\mathrm{the}\:\mathrm{acceleration} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{train}\:\mathrm{has}\:\mathrm{no}\:\mathrm{influence}\:\mathrm{on}\:\mathrm{the} \\ $$$$\mathrm{stone}. \\ $$
Commented by Tinkutara last updated on 14/Jun/17
$$\mathrm{Thanks}\:\mathrm{Sir}! \\ $$
Commented by ajfour last updated on 14/Jun/17
$${its}\:{initial}\:{velocity}\:{is}\:{equal}\:{to}\: \\ $$$${that}\:{of}\:{the}\:{train}\:{when}\:{it}\:{was} \\ $$$${released}\:{from}\:{passenger}'{s}\:{hand} \\ $$$${so}\:{it}\:{will}\:{move}\:{both}\:{forward}\:{and} \\ $$$${downwards}\:{along}\:{a}\:{parabola}.. \\ $$