Distance travelled by a projectile in 2d When a projectile is projected in 2d is there a way to find the dist{the actual path lenghth} travelled by the projectlie in a a given time ? Can we do it by eavaluating the radius vector at times t and t+dt and then subtracting the former from  the latter. For small choices of dt the the displacement vector so obtained would be approximately equal to the actual path. Then should we integrate the function obtained {this part is a bit unclear} from t1 to t2 to get a dist function. Is there a simpler way to approach this problem?
 A: What you need to calculate is what's known as the Arc Length. 
So, you have the parametric equations:
$$x = u_xt$$
$$y = u_yt - \frac{g}{2}t^2$$
Where $u_x$ is the horizontal starting velocity, $u_y$ is the vertical starting velocity, $t$ is the time from launch and $g$ is the acceleration due to gravity ($9.81$ on Earth).
Now, using the formula here, we can start to find the arc length, $L$ between the times $\alpha$ and $\beta$:
$$L = \int^\beta_\alpha \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} dt$$
Next, we differentiate the equation:
$$\frac{dx}{dt} = u_x$$
$$\frac{dy}{dt} = u_y - gt$$
And insert it into the equation:
$$L = \int^\beta_\alpha \sqrt{u_x^2 + \left(u_y - gt\right)^2} dt$$
$$L = \int^\beta_\alpha \sqrt{u_x^2 + u_y^2 - 2u_ygt + g^2t^2} dt$$
Note that $u_x^2 + u_y^2 = u^2$, where $u$ is the starting velocity (in the direction of the launch), so:
$$L = \int^\beta_\alpha \sqrt{u^2 - 2u_ygt + g^2t^2} dt$$
Next, we need to find $\alpha$ and $\beta$ by setting $y = 0$ (assuming the projectile is launched from and lands on the ground):
$$0 = t\left(u_y-\frac{g}{2}t\right)$$
Therefore,
$$\alpha = 0$$
$$\beta = \frac{2u_y}{g}$$
You should be able to solve the integral now to get the path length. You should probably use numerical integration for this.
A: It is easier to find modulus of velocity vector (speed) as function of time and intergrate wrt to dt to get distance
