The equations of motion of a projectile are given by x=36t ~m and 2y=96t9.8t^2 ~m. The angle of projection is A arcsin(4/5) B arcsin(3/5) C arcsin(4/3) D arcsin(3/4)
Question
The equations of motion of a projectile are given by $x=36t \mathrm{~m}$ and $2y=96t9.8t^{2} \mathrm{~m}$. The angle of projection is
A $\sin^{1}\left(\frac{4}{5}\right)$ B $\sin^{1}\left(\frac{3}{5}\right)$
C $\sin^{1}\left(\frac{4}{3}\right)$ D $\sin^{1}\left(\frac{3}{4}\right)$
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Answer
To determine the angle of projection for the given equations of motion for a projectile:

The equations are:
 ( x = 36t ) (horizontal motion)
 ( 2y = 96t  9.8t^2 ) (vertical motion)

Simplify the vertical motion equation to:
 ( y = 48t  4.9t^2 )

Let ( u ) be the initial velocity and ( \theta ) be the angle of projection.
Horizontal Component
 The horizontal component of the velocity, ( u_x ), is: $$ u \cos \theta = \frac{dx}{dt} \text{ at } t = 0 $$
 Given ( x = 36t ): $$ \frac{dx}{dt} = 36 $$
 So, $$ u \cos \theta = 36 \quad \text{(Equation 1)} $$
Vertical Component
 The vertical component of the velocity, ( u_y ), is: $$ u \sin \theta = \frac{dy}{dt} \text{ at } t = 0 $$
 Given ( y = 48t  4.9t^2 ): $$ \frac{dy}{dt} = 48  9.8t $$
 At ( t = 0 ): $$ \frac{dy}{dt} = 48 $$
 So, $$ u \sin \theta = 48 \quad \text{(Equation 2)} $$
Determining the Angle

To find ( \theta ), divide Equation 2 by Equation 1: $$ \tan \theta = \frac{u \sin \theta}{u \cos \theta} = \frac{48}{36} = \frac{4}{3} $$

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

Using a right triangle with sides corresponding to the ratio ( 4:3:5 ) (Pythagorean triplet):
 Opposite (perpendicular) = 4
 Adjacent (base) = 3
 Hypotenuse = 5

Therefore, ( \sin \theta ) is: $$ \sin \theta = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{4}{5} $$

The angle of projection, ( \theta ), is: $$ \theta = \sin^{1}\left(\frac{4}{5}\right) $$
Final Answer
So the correct option is:
A ( \sin^{1}\left( \frac{4}{5} \right) )
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