Kinematics with Force Field

When $x <15$ the equations of motion are:

$\ddot{x} = 0$ $\ddot{y} = -10$

$\dot{x} = \frac{20}{\sqrt{2}}$ $\dot{y }= \frac{20}{\sqrt{2}} - 10t$

$x = \frac{20t}{\sqrt{2}}$ $y = \frac{20t}{\sqrt{2}} - 5t^2$

Replacing $t$ with $x$ in $y$ :

$y = x = \frac{x^2}{40}$

So, at $x=15$ , $y=9.375$ . When $15 \le x \le 25$ , the particle experiences no force and therefore moves in a straight line. The slope of this straight line is determined by the velocity components of the particle at $x=15$ which on solving are:

$v_x = \frac{20}{\sqrt{2}}$ $v_y = \frac{5}{\sqrt{2}}$

Therefore, the slope is:

$\frac{dy}{dx} = 0.25$ $y = 0.25x + c$

Using the boundary condition that at $x=15$ , $y=9.375$ , $c=5.625$ . Therefore, at $x=25$ the $y$ coordinate of the particle is $y=11.875$ .

At this instant, which is now considered for convenience as $t=0$ , we have:

$\dot{x}(0) = \frac{20}{\sqrt{2}}$ $\dot{y}(0) = \frac{5}{\sqrt{2}}$ $x(0) = 25$ $y(0) = 11.875$

$\ddot{x} = 0$ $\ddot{y} = -10$

Solving for $x$ and $y$ , applying initial conditions and writing $y$ in terms of $x$ gives:

$y = \frac{x-25}{4} - \frac{(x-25)^2}{40} + 11.875$

Finally, when $y=0$ , then: $0 = \frac{x-25}{4} - \frac{(x-25)^2}{40} + 11.875$

Let $x-25=z$ , then $0 = \frac{z}{4} - \frac{z^2}{40} + 11.875$

Solving this quadratic equation leads to $\boxed{x=52.3607 \ \mathrm{m}}$ .

I also performed a simulation using a computer program and the trajectory of the particle is as follows: