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**Hint:**The vertical and horizontal motions are independent of each other. Differentiate the equation of trajectory of a profile with respect to $ x $ and equate the result to zero. Find the value of $ x $ and compute it in the equation of trajectory of a profile.

**Formula Used:**The formulae used in the solution are given here.

$ y = x\tan \theta $ where $ y $ is the vertical distance, $ x $ is the horizontal distance and the angle of projection is $ \theta $ .

**Complete step by step solution:**

Projectile motion is the motion of an object thrown or projected into the air, subject to only the acceleration of gravity. The object is called a projectile, and its path is called its trajectory. The motions along perpendicular axes are independent and thus can be analysed separately.

Then, resolve the position and/or velocity of the object in the horizontal and vertical components. The components of position s are given by the quantities $ x $ and $ y $ , and the components of the velocity v are given by $ {v_x} = v\cos \theta $ and $ {v_y} = v\sin \theta $ , where $ v $ is the magnitude of the velocity and $ \theta $ is its direction.

For horizontal motion, we have acceleration along the x-axis, $ {a_x} = 0 $ .

Using these values in the equations, of motion, we get,

$ x = {x_0} + {v_x}t $

$ y = ax - b{x^2} $

It has been given that, the equation of trajectory of a profile is $ y = bx - a{x^2} $ where symbols have their usual meanings and $ a $ and $ b $ are positive constants.

We differentiate this equation of projectile with respect to $ x $ . We get,

$ \dfrac{{dy}}{{dx}} = b - 2ax $ .

Thus, the angle of horizontal projection $ \theta $ is given by,

$ \tan \theta = \dfrac{{dy}}{{dx}} = a - 2bx $

Now when we consider, $ \dfrac{{dy}}{{dx}} = 0 $ ,

We have $ x = \dfrac{b}{{2a}} $ .

Substituting $ x = \dfrac{b}{{2a}} $ in the equation of trajectory of a profile, we have,

$ y = b\left( {\dfrac{b}{{2a}}} \right) - a{\left( {\dfrac{b}{{2a}}} \right)^2} $

$ \Rightarrow y = \dfrac{{{a^2}}}{{2b}} - \dfrac{{{a^2}}}{{4b}} = \dfrac{{{a^2}}}{{4b}} $

Put, $ x = 0 $ in the equation, $ \tan \theta = a - 2bx $ . Thus we get

$ \tan \theta = a $

$ \Rightarrow \theta = {\tan ^{ - 1}}\left( a \right) $

The angle of projection (from the horizontal) for the projectile is $ {\tan ^{ - 1}}\left( a \right) $ .

**Hence the correct answer is Option B.**

**Note:**

When we speak of the range of a projectile on level ground, we assume that range is very small compared with the circumference of the Earth. If, however, the range of the projectile is large, the Earth curves away below the projectile and acceleration of gravity changes direction along the path. The range is larger than predicted by the range equation given above because the projectile has farther to fall than it would on level ground. If the initial speed is great enough, the projectile goes into orbit. This is called escape velocity. This possibility was recognized centuries before it could be accomplished. When an object is in orbit, the Earth curves away from underneath the object at the same rate as it falls. The object thus falls continuously but never hits the surface. These and other aspects of orbital motion, such as the rotation of the Earth, will be covered analytically and in greater depth later in this text. Once again we see that thinking about one topic, such as the range of a projectile, can lead us to others, such as the Earth orbits.

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