Question Number 31148 by NECx last updated on 03/Mar/18
![A body is projected at an angle 35° with an initial speed of 45m/s. What is the velocity of the body after 2s and the angle made with the horizontal axis? [g=9.81ms^(−2) ]](https://www.tinkutara.com/question/Q31148.png)
$${A}\:{body}\:{is}\:{projected}\:{at}\:{an}\:{angle} \\ $$$$\mathrm{35}°\:{with}\:{an}\:{initial}\:{speed}\:{of}\:\mathrm{45}{m}/{s}. \\ $$$${What}\:{is}\:{the}\:{velocity}\:{of}\:{the}\:{body} \\ $$$${after}\:\mathrm{2}{s}\:{and}\:{the}\:{angle}\:{made}\:{with} \\ $$$${the}\:{horizontal}\:{axis}?\:\left[{g}=\mathrm{9}.\mathrm{81}{ms}^{−\mathrm{2}} \right] \\ $$
Commented by NECx last updated on 03/Mar/18
$${for}\:{the}\:{velocity}\:{in}\:{time}\:{t},\:{we}\:{have} \\ $$$$\:{V}\left({t}\right)={u}\mathrm{sin}\:\theta−{gt} \\ $$$${v}\left(\mathrm{2}\right)=\mathrm{45}{sin}\mathrm{35}\:−\:\mathrm{9}.\mathrm{81}×\mathrm{2} \\ $$$$\:\:\:\:=\mathrm{25}.\mathrm{81094}−\mathrm{19}.\mathrm{62} \\ $$$$\:\:\:\:\:=\mathrm{6}.\mathrm{19}{m}/{s} \\ $$$$ \\ $$$${then}\:{I}\:{got}\:{confused}\:{on}\:{seeing}\:{angle} \\ $$$${made}\:{with}\:{the}\:{horizontal}.{Please} \\ $$$${help}.\:{Thanks} \\ $$
Answered by mrW2 last updated on 03/Mar/18
$${u}=\mathrm{45}\:{m}/{s} \\ $$$${u}_{{x}} =\mathrm{45}\:\mathrm{cos}\:\mathrm{35}° \\ $$$${u}_{{y}} =\mathrm{45}\:\mathrm{sin}\:\mathrm{35}° \\ $$$${v}_{{x}} \left({t}\right)={u}_{{x}} \\ $$$${v}_{{y}} \left({t}\right)={u}_{{y}} −{gt} \\ $$$${v}\left({t}\right)=\sqrt{{v}_{{x}} ^{\mathrm{2}} +{v}_{{y}} ^{\mathrm{2}} } \\ $$$$\mathrm{tan}\:\theta\left({t}\right)=\frac{{v}_{{y}} }{{v}_{{x}} } \\ $$$$\theta\left({t}\right)=\mathrm{tan}^{−\mathrm{1}} \left(\frac{{v}_{{y}} }{{v}_{{x}} }\right) \\ $$$$ \\ $$$${at}\:{t}=\mathrm{2}\:{s}: \\ $$$${v}_{{x}} \left(\mathrm{2}\right)=\mathrm{45}\:\mathrm{cos}\:\mathrm{35}°=\mathrm{36}.\mathrm{86}\:{m}/{s} \\ $$$${v}_{{y}} \left(\mathrm{2}\right)=\mathrm{45}\:\mathrm{sin}\:\mathrm{35}°−\mathrm{9}.\mathrm{81}×\mathrm{2}=\mathrm{6}.\mathrm{19}\:{m}/{s} \\ $$$${v}\left(\mathrm{2}\right)=\sqrt{\mathrm{36}.\mathrm{86}^{\mathrm{2}} +\mathrm{6}.\mathrm{19}^{\mathrm{2}} }=\mathrm{37}.\mathrm{4}\:{m}/{s} \\ $$$$\theta\left(\mathrm{2}\right)=\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{6}.\mathrm{19}}{\mathrm{36}.\mathrm{86}}=\mathrm{9}.\mathrm{5}° \\ $$
Commented by NECx last updated on 03/Mar/18
$${thank}\:{you}\:{so}\:{much}\:{sir} \\ $$