How Do You Calculate the Landing Point of a Projectile Thrown from a Height?

[Darth Frodo]

Homework Statement

A stone is thrown with a velocity v_{0} at an angle \alpha to the horizontal (see image) from a step of height H. Calculate the x coordinate x_{1} of the point where the stone hits the ground.

Homework Equations

x(t) = u_{x}t + x_{0}
y(t) = u_{y}t + \frac{1}{2}at^{2} + y_{0}

The Attempt at a Solution

u_{x} = v_{0}cos(\alpha)
u_{y} = -v_{0}sin(\alpha)

x_{1} = u_{x}t_{1}
x_{1} = v_{0}cos(\alpha)t_{1}
y(t) = 0 @ t_{1}
0 = -v_{0}sin(\alpha)t_{1} - \frac{g}{2}(t_{1})^2 + H


Here is where I have a discrepency with the provided (Bare Bones) Solution.

According to the solution; 0 = v_{0}sin(\alpha)t_{1} - \frac{g}{2}(t_{1})^2 + H

This seems to be missing the "-" in front of the 0 = v_{0}sin(\alpha)t_{1} term.

My question is am I correct?

 

[Tallus Bryne]

You could say that both your result and the solution are correct; they just make different assumptions about how the angle α is determined.

Your result assumes that α is a positive angle which lies between the positive x-axis and the initial velocity vector.

For the given solution to be correct, it is assumed that the angle α must be determined by rotating counter-clockwise from the positive x-axis for positive angles (or clockwise for negative angles).

 

[Gzyousikai]

x(t)=v_0\cos\alpha t, y(t)=H+v_0\sin\alpha t-\frac{1}{2}gt^2
combine those two parametric equations, we get
y=x\tan\alpha-\frac{g\sec^2\alpha}{2v_0^2}x^2+H

let y=0 and solve for x