A body is projected from a plane surface at an angle theta. Find the value of theta at which the parabolic path travelled by the body is maximum.And also find the velocity with which it should be projected.
This is a physics problem. 1. The only problem is that it doesn’t have physics. Only maths. 2. The problem uses some calculus , (integration,mostly) First broke the x and y direction velocity and integrated it with w.r.t time. But I couldn’t proceed further because a relation came in logarithm with base e.
So please can anyone help.
Edit: Removed classical physics tag.
This is a standard problem. Assuming standard conditions (gravity being a constant, in the -ve y direction), we can write the velocity of the object as a function of time as
$$v(t) = v_0\cos\theta \textbf{i} + (v_o\sin\theta - gt)\textbf{j}$$
Now, the distance travelled along the trajectory at a particular time instant is given as
$$ds = \sqrt{dx^2 + dy^2} \\ = \sqrt{(v_0\cos\theta)^2 + (v_0\sin\theta - gt)^2}dt$$
Hence, the total length traversed in the air is given by
$$S = \int_0^{t_f}\sqrt{(v_0\cos\theta)^2 + (v_0\sin\theta - gt)^2}dt$$
Now, this is a function of both the initial velocity, and the angle of projection. Hence, to maximise it, we need to set the partial derivative w.r.t angle of projection to 0
$$\frac{\partial S}{\partial \theta} = \frac{\partial t_f}{\partial \theta}.{}\int_0^{t_f}\frac{-2gt\cos\theta dt}{2\sqrt{(v_0\cos\theta)^2 + (v_0\sin\theta - gt)^2}} = 0$$
Now, time of flight is given from the other kinematic equation, as
$$t_f = \sqrt{2v_0g\sin\theta} \implies \frac{\partial t_f}{\partial\theta} = \frac{v_og\cos\theta}{\sqrt{2v_0g\sin\theta}}$$
Put it all together and can you proceed?