Question Number 55418 by Tawa1 last updated on 23/Feb/19
Commented by tanmay.chaudhury50@gmail.com last updated on 23/Feb/19
$${pls}\:{calculate}\:{sin}\mathrm{2}\theta=\frac{\mathrm{5}}{\mathrm{13}}\:{using}\:{calculatr} \\ $$$${i}\:{have}\:{not}\:{downloaded}\:{scientific}\:{calcuator}.. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 23/Feb/19
![Range=vcosα×2t_o 0=vsinα−gt_0 t_0 =((vsinα)/g) Range=vcosα×2×((vsinα)/g)=((v^2 sin2α)/g) (Range)_(max) =(v^2 /g)×1 [max value of sin2α=1] (Range)_(max) =(v^2 /g) as per question (Range)_(max) =R=(v^2 /g)](https://www.tinkutara.com/question/Q55420.png)
$${Range}={vcos}\alpha×\mathrm{2}{t}_{{o}} \\ $$$$\mathrm{0}={vsin}\alpha−{gt}_{\mathrm{0}} \\ $$$${t}_{\mathrm{0}} =\frac{{vsin}\alpha}{{g}} \\ $$$${Range}={vcos}\alpha×\mathrm{2}×\frac{{vsin}\alpha}{{g}}=\frac{{v}^{\mathrm{2}} {sin}\mathrm{2}\alpha}{{g}} \\ $$$$\left({Range}\right)_{{max}} =\frac{{v}^{\mathrm{2}} }{{g}}×\mathrm{1}\:\:\left[{max}\:{value}\:{of}\:{sin}\mathrm{2}\alpha=\mathrm{1}\right] \\ $$$$\left({Range}\right)_{{max}} =\frac{{v}^{\mathrm{2}} }{{g}} \\ $$$${as}\:{per}\:{question}\:\left({Range}\right)_{{max}} ={R}=\frac{{v}^{\mathrm{2}} }{{g}} \\ $$
Commented by Tawa1 last updated on 24/Feb/19
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 23/Feb/19
$${Range}=\frac{\mathrm{5}{R}}{\mathrm{13}}=\frac{\mathrm{5}}{\mathrm{13}}×\frac{{v}^{\mathrm{2}} }{{g}} \\ $$$${now}\:{range}=\frac{{v}^{\mathrm{2}} {sin}\mathrm{2}\theta}{{g}}=\frac{\mathrm{5}{v}^{\mathrm{2}} }{\mathrm{13}{g}} \\ $$$${sin}\mathrm{2}\theta=\frac{\mathrm{5}}{\mathrm{13}} \\ $$$$\theta=\frac{\mathrm{1}}{\mathrm{2}}{sin}^{−\mathrm{1}} \left(\frac{\mathrm{5}}{\mathrm{13}}\right)\:{and}\:\pi−\frac{\mathrm{1}}{\mathrm{2}}{sin}^{−\mathrm{1}} \left(\frac{\mathrm{5}}{\mathrm{13}}\right) \\ $$$$ \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 23/Feb/19
![time of flight=2t_0 =((2vsinθ)/g) [0=vsinθ−gt_0 ] now sin2θ=(5/(13)) θ=(1/2)sin^(−1) ((5/(13))) and π−(1/2)sin^(−1) ((5/(13))) so time of flight ... (time)_(flight) = ((2vsin((1/2)sin^(−1) (5/(13))))/(2g)) and ((2vsin[π−(1/2)sin^(−1) ((5/(13)))])/(2g))](https://www.tinkutara.com/question/Q55426.png)
$${time}\:{of}\:{flight}=\mathrm{2}{t}_{\mathrm{0}} \:=\frac{\mathrm{2}{vsin}\theta}{{g}} \\ $$$$ \\ $$$$\left[\mathrm{0}={vsin}\theta−{gt}_{\mathrm{0}} \right] \\ $$$${now}\:{sin}\mathrm{2}\theta=\frac{\mathrm{5}}{\mathrm{13}} \\ $$$$\theta=\frac{\mathrm{1}}{\mathrm{2}}{sin}^{−\mathrm{1}} \left(\frac{\mathrm{5}}{\mathrm{13}}\right)\:{and}\:\pi−\frac{\mathrm{1}}{\mathrm{2}}{sin}^{−\mathrm{1}} \left(\frac{\mathrm{5}}{\mathrm{13}}\right) \\ $$$${so}\:{time}\:{of}\:{flight}\:… \\ $$$$ \\ $$$$\left({time}\right)_{{flight}} =\:\frac{\mathrm{2}{vsin}\left(\frac{\mathrm{1}}{\mathrm{2}}{sin}^{−\mathrm{1}} \frac{\mathrm{5}}{\mathrm{13}}\right)}{\mathrm{2}{g}} \\ $$$${and}\:\frac{\mathrm{2}{vsin}\left[\pi−\frac{\mathrm{1}}{\mathrm{2}}{sin}^{−\mathrm{1}} \left(\frac{\mathrm{5}}{\mathrm{13}}\right)\right]}{\mathrm{2}{g}} \\ $$$$ \\ $$
Commented by malwaan last updated on 24/Feb/19
$$\mathrm{sin}^{−\mathrm{1}} \left(\frac{\mathrm{5}}{\mathrm{13}}\right)\approx\mathrm{22}.\mathrm{62} \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 24/Feb/19
$${thank}\:{you}\:{sir}… \\ $$