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The equations of motion of a projectile are given by $x=36t$ and $2y=96t-9.8{{t}^{2}}$ . The angle of projection is
Option: 1
${{\sin }^{-1}}(\frac{4}{5})$

Option: 2
${{\sin }^{-1}}(\frac{3}{5})$

Option: 3
${{\sin }^{-1}}(\frac{4}{3})$

Option: 4
${{\sin }^{-1}}(\frac{3}{4})$

Answers (1)

Answer : A) ${{\sin }^{-1}}(\frac{4}{5})$

Explanation : Given,

$\[x=36t\] \\ 2y=96t-9.8{{t}^{2}}$

Or $y=48t-4.9{{t}^{2}}$

Then the initial horizontal component of velocity is,

${{u}_{x}}=u\cos \theta$

${{u}_{x}}=36$

$u\cos \theta =36......(1)$

Similarly, the initial vertical component of velocity is,

${{u}_{y}}=u\sin \theta$

${{u}_{y}}=48$

$u\sin \theta =48 ......(2)$

Dividing equation (2) by (1),

$\tan \theta =\frac{48}{36}$

$\tan \theta =\frac{4}{3}$

$\therefore \sin \theta =\frac{4}{5}$

$\theta ={{\sin }^{-1}}(\frac{4}{5})$

Posted by

Sumit Saini

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