The equations of motion of a projectile are given by $x=36 t$ metre and $2 y=96 t-9.8 t^2$ metre. The angle…
The equations of motion of a projectile are given by $x=36 t$ metre and $2 y=96 t-9.8 t^2$ metre. The angle of projection is :
- $\sin ^{-1}\left(\frac{4}{5}\right)$
- $\sin ^{-1}\left(\frac{3}{5}\right)$
- $\sin ^{-1}\left(\frac{4}{3}\right)$
- $\sin ^{-1}\left(\frac{3}{4}\right)$
Solution
$x=36 t$ metre
$2 y=\left(96 t-9.8 t^2\right)$ metre
Standard equations are,
$x=(u \cos \theta) t$
$y=u \sin \theta t-\frac{1}{2} g t^2$
$u \cos =36$
$u \sin \theta=\frac{96}{2}=48$
$\therefore \quad \tan \theta=\frac{u \sin \theta}{u \cos \theta}=\frac{48}{36}=\frac{4}{3}$
$\theta=\tan ^{-1} \frac{4}{3}$
or $\theta=\sin ^{-1} \frac{4}{5}$
Asked in: AP EAMCET 2003
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