# What is the horizontal range of the stone when the stone strikes the ground?

## A person is standing on top of a hill that slopes downwards uniformly at an angle ! with respect to the horizontal. The person throws a stone at an initial angle !0 from the horizontal with an initial speed of v0 . You may neglect air resistance. What is the horizontal range of the stone when the stone strikes the ground? Can I set the coordinate system as follows: Choose the origin to be at the tip of the inclined plane, the x-axis along the inclined plane and the y-axis perpendicular to the inclined plane. With this coordinate system, the angle the initial speed makes with the x-axis is θ+Φ. We resolve the acceleration due to gravity and initial velocity vector into their 2 components ax+ay=(gsinΦ)i-(gcosΦ)j v0=vxi+Vyj=v0cos(θ+Φ)i+v0sin(θ+Φ)j Then write down the complete set of kinematics for each independent direction x and y describing the position and velocity at instant, t. Why is my approach to solving the problem incorrect?

Oct 6, 2017

I can answer in principle, but not give a specific answer as the data hasn't appeared in the question.

#### Explanation:

Firstly, make sure you analyse vertical and horizontal motion separately - they are completely independent.

You need to find the time of flight (suggest $s = u t + \frac{1}{2} a {t}^{2}$) just looking at the trajectory in the vertical plane and then, as there is no air resistance, realise that the horizontal velocity is constant, so just use $v = \frac{s}{t}$ to find the horizontal displacement.

Oct 11, 2017

Yes, you could set the coordinate system as you suggest; and then convert from down the slope distance to horizontal distance for the final answer.

#### Explanation:

If you did what you suggested, and it was marked incorrect, I suspect you did not then convert down the slope distance to horizontal distance. If the distance down the slope you calculated was x, then the horizontal range, ${s}_{h}$ would be
${s}_{h} = x \cdot \cos \phi$

I hope this helps,
Steve

Oct 11, 2017

See below.

#### Explanation:

Assuming the origin of coordinates at the very top, the kinematic movement equations are

$\left(x , y\right) = \left({v}_{0} \cos \theta t , {v}_{0} \sin \theta t - \frac{1}{2} g {t}^{2}\right)$

The inclined plane is described by

$\left(x , y\right) = d \left(\cos \phi , - \sin \phi\right)$

so the intersection is given at

$\left\{\begin{matrix}\left({v}_{0} \cos \theta\right) t = d \cos \phi \\ \left({v}_{0} \sin \theta\right) t - \frac{1}{2} g {t}^{2} = - d \sin \phi\end{matrix}\right.$

now solving for $t , d$ we obtain

{(t =(2 sin(theta+phi)v_0)/(cosphi g)), (d = ( 2 costheta sin(theta+phi) v_0^2)/(cos^2phi g)):}