Travel distance of a particle

Question

How can we show that a projectile fired at an angle $\theta$ with initial speed $v_0$ travels a total distance $\frac{v_0^2}{g}\sin2\theta$ before hitting the ground?
The way I set it up is: direction is $(\cos\theta,\sin\theta)$. By using Newton's law, $$r(t) = (0,-\frac{1}{2} gt^2)+tv_0(\cos\theta,\sin\theta) = \frac{v_0^2}{g}\sin2\theta$$ I tried to solve the LHS so that it matches the RHS, the total distance. But obviously, there is something wrong with the way I tried to approach it. But cannot see it.

Any hints and pointers in terms of setting it up would be great. Also I do not want the full solution but rather few small hints to be able to set it up correctly.

Answer 1

Hint: Find out how long it takes for the $y$-component of velocity to reach $0$. Call that time $t_0$.

By symmetry it takes time $2t_0$ to hit the ground.

Answer 2

If you have learned differential equation, you can represent the displacement ${\bf r}(t) = (r_1,r_2)$ by an initial value problem: $${\bf r}''(t) = (0,-g) \\ {\bf r} (0) = (0,0) \\ {\bf r}'(0) = {\bf v}(0) = (v_0 \cos(\theta), v_0 \sin(\theta)) $$ solve the y component = zero for a time, then plugging it in the x component. This approach I learned in ODE class is good, for air friction fits perfectly for the model, gravitatiional constant change too : $$ {\bf r}''(t) = (0,-\frac{GM}{(r_2+R)^2}) - \gamma {\bf r}'(t) $$ $M$ is the mass of the earth and $R$ is the radius.