Menu Close

Question-52818




Question Number 52818 by Tawa1 last updated on 13/Jan/19
Answered by tanmay.chaudhury50@gmail.com last updated on 13/Jan/19
V_(cart) =4i  (V^ )_(cart) ^(stone) =velocity of stone w.r.t man(cart)  V_(cart) ^(stone) =V_(earth) ^(stone) −V_(earth) ^(cart)   6cos30^o k+6sin30^0 j+4i=V_(earth) ^(stone)   m_(stone) ×V_(earth) ^(stone) =(m_(stone) +m_(object) )V_(earth) ^(combined)   at highest point V_z  component=0 for stone  V_(earth) ^(combined)  at highest point=((0×k+6sin30^o j+4i)/2)  =((0×k+3j+4i)/2)=2i+(3/2)j+0×k  speed combined=∣V_(earth) ^(combined) ∣=(√(2^2 +((3/2))^2 )) =(√(4+2.25)) =2.5m/sec      b)v=(√(gl))   l=(v^2 /g)=((6.25)/(9.8))=0.64meter
$${V}_{{cart}} =\mathrm{4}{i} \\ $$$$\left({V}^{} \right)_{{cart}} ^{{stone}} ={velocity}\:{of}\:{stone}\:{w}.{r}.{t}\:{man}\left({cart}\right) \\ $$$${V}_{{cart}} ^{{stone}} ={V}_{{earth}} ^{{stone}} −{V}_{{earth}} ^{{cart}} \\ $$$$\mathrm{6}{cos}\mathrm{30}^{{o}} {k}+\mathrm{6}{sin}\mathrm{30}^{\mathrm{0}} {j}+\mathrm{4}{i}={V}_{{earth}} ^{{stone}} \\ $$$${m}_{{stone}} ×{V}_{{earth}} ^{{stone}} =\left({m}_{{stone}} +{m}_{{object}} \right){V}_{{earth}} ^{{combined}} \\ $$$${at}\:{highest}\:{point}\:{V}_{{z}} \:{component}=\mathrm{0}\:{for}\:{stone} \\ $$$${V}_{{earth}} ^{{combined}} \:{at}\:{highest}\:{point}=\frac{\mathrm{0}×{k}+\mathrm{6}{sin}\mathrm{30}^{{o}} {j}+\mathrm{4}{i}}{\mathrm{2}} \\ $$$$=\frac{\mathrm{0}×{k}+\mathrm{3}{j}+\mathrm{4}{i}}{\mathrm{2}}=\mathrm{2}{i}+\frac{\mathrm{3}}{\mathrm{2}}{j}+\mathrm{0}×{k} \\ $$$${speed}\:{combined}=\mid{V}_{{earth}} ^{{combined}} \mid=\sqrt{\mathrm{2}^{\mathrm{2}} +\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} }\:=\sqrt{\mathrm{4}+\mathrm{2}.\mathrm{25}}\:=\mathrm{2}.\mathrm{5}{m}/{sec} \\ $$$$ \\ $$$$ \\ $$$$\left.{b}\right){v}=\sqrt{{gl}}\:\:\:{l}=\frac{{v}^{\mathrm{2}} }{{g}}=\frac{\mathrm{6}.\mathrm{25}}{\mathrm{9}.\mathrm{8}}=\mathrm{0}.\mathrm{64}{meter} \\ $$
Commented by Tawa1 last updated on 13/Jan/19
God bless you sir
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 13/Jan/19
is my answer correct...
$${is}\:{my}\:{answer}\:{correct}… \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *