Question

The equations of motion of a projectile are given by $x=36t \mathrm{~m}$ and $2y=96t-9.8t^{2} \mathrm{~m}$. 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:

  1. The equations are:

    • ( x = 36t ) (horizontal motion)
    • ( 2y = 96t - 9.8t^2 ) (vertical motion)
  2. Simplify the vertical motion equation to:

    • ( y = 48t - 4.9t^2 )
  3. Let ( u ) be the initial velocity and ( \theta ) be the angle of projection.

Horizontal Component

  1. The horizontal component of the velocity, ( u_x ), is: $$ u \cos \theta = \frac{dx}{dt} \text{ at } t = 0 $$
  2. Given ( x = 36t ): $$ \frac{dx}{dt} = 36 $$
  3. So, $$ u \cos \theta = 36 \quad \text{(Equation 1)} $$

Vertical Component

  1. The vertical component of the velocity, ( u_y ), is: $$ u \sin \theta = \frac{dy}{dt} \text{ at } t = 0 $$
  2. Given ( y = 48t - 4.9t^2 ): $$ \frac{dy}{dt} = 48 - 9.8t $$
  3. At ( t = 0 ): $$ \frac{dy}{dt} = 48 $$
  4. So, $$ u \sin \theta = 48 \quad \text{(Equation 2)} $$

Determining the Angle

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

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

  3. Using a right triangle with sides corresponding to the ratio ( 4:3:5 ) (Pythagorean triplet):

    • Opposite (perpendicular) = 4
    • Adjacent (base) = 3
    • Hypotenuse = 5
  4. Therefore, ( \sin \theta ) is: $$ \sin \theta = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{4}{5} $$

  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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