Question Number 70669 by jagannath19 last updated on 06/Oct/19
Commented by jagannath19 last updated on 06/Oct/19
$${answer}\:{with}\:{explanation}\:{please} \\ $$
Commented by mr W last updated on 06/Oct/19
$${a}_{{R}} ={g}\:\mathrm{cos}\:\varphi \\ $$$$\varphi\:{varies}\:{from}\:\theta\:{to}\:{zero}\:{and}\:{from}\:{zero} \\ $$$${to}\:\theta,\:{therefore}\:\left({a}\right)\:{or}\:\left({c}\right)\:{could}\:{be}\:{true}, \\ $$$${but}\:{the}\:{variation}\:{of}\:\mathrm{tan}\:\varphi\:{is}\:{linear}, \\ $$$${and}\:{the}\:{variation}\:{of}\:\mathrm{cos}\:\varphi\:{is}\:{not} \\ $$$${linear},\:{so}\:{the}\:{only}\:{correct}\:{answer}\:{is} \\ $$$$\left({a}\right). \\ $$
Commented by mr W last updated on 06/Oct/19
$${if}\:{you}\:{want}\:{to}\:{get}\:{the}\:{exact}\:{graph}\:{from} \\ $$$${a}_{{R}} \:{with}\:{respect}\:{to}\:{t}: \\ $$$${x}={v}_{\mathrm{0}} \:\mathrm{cos}\:\theta\:{t} \\ $$$${y}={v}_{\mathrm{0}} \:\mathrm{sin}\:\theta\:{t}−\frac{\mathrm{1}}{\mathrm{2}}{gt}^{\mathrm{2}} \\ $$$$\mathrm{tan}\:\varphi=\frac{{dy}}{{dx}}=\frac{\frac{{dy}}{{dt}}}{\frac{{dx}}{{dt}}}=\frac{{v}_{\mathrm{0}} \mathrm{sin}\:\theta−{gt}}{{v}_{\mathrm{0}} \mathrm{cos}\:\theta}=\mathrm{tan}\:\theta−\frac{{gt}}{{v}_{\mathrm{0}} \:\mathrm{cos}\:\theta} \\ $$$$\mathrm{cos}\:\varphi=\frac{\mathrm{1}}{\sqrt{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\varphi}}=\frac{\mathrm{1}}{\sqrt{\mathrm{1}+\left(\mathrm{tan}\:\theta−\frac{{gt}}{{v}_{\mathrm{0}} \mathrm{cos}\:\theta}\right)^{\mathrm{2}} }} \\ $$$${a}_{{R}} ={g}\:\mathrm{cos}\:\varphi \\ $$$${a}_{{R}} =\frac{{g}}{\sqrt{\mathrm{1}+\left(\mathrm{tan}\:\theta−\frac{{gt}}{{v}_{\mathrm{0}} \mathrm{cos}\:\theta}\right)^{\mathrm{2}} }} \\ $$$${the}\:{graph}\:{of}\:{a}_{{R}} \:{w}.{r}.{t}.\:{t}\:{see}\:{following} \\ $$$${example}: \\ $$
Commented by mr W last updated on 06/Oct/19
Commented by jagannath19 last updated on 07/Oct/19
$${thank}\:{you}\:{so}\:{much}\:{sir} \\ $$