# How does moving the launch area forward impact the area of impact?

If we launch an object in some way, and it lands at distance x, if we move forward the launching mechanism y units forward or backwards, will the landing distance also move forwards or backwards y units? Assume the launching force is the same.

Here's an example to make it easier to understand. If we throw a ball one foot away from us, and then if we move 6 inches forward, if we throw the ball again, will it land 6 inches forward from where it initially landed? If not, is there a formula relating the distance you move and the distance the projectile will move?

• It would be rather odd if the answer to "How far can you throw?" was "It depends on where I'm standing". There's no mathematical reason why you should be able to throw a ball farther in your backyard compared to your front yard, or vice versa. Commented Mar 11 at 18:51

For example, a launched projectile experiences only the gravitational force, which under suitable circumstances is constant, i.e. the acceleration due to gravity is $$g=9.81\text{m}\text{s}^2$$. So we have: $$x(t)={1\over 2}gt^2+v_0t+x_0,$$ for the horizontal component of the motion. Not that the solution depends on the initial condition of the problem, i.e. the initial velocity $$v_0$$ and initial position $$x_0$$. The $$x_0$$ term expresses exactly your concern, e.g. if you were initially at some point, call it "ground zero" where we take $$x_0=0$$, then the object moves horizontally a distance $$x(t)$$ in distance during the time from $$t_0=0,t$$: $$\text{distance}_1=x(t)={1\over 2}gt^2+v_0t.$$ Now consider the same problem except your new launching point is $$20\text{m}$$ forward of your original starting point so that $$x_0=20$$, you have: $$\text{distance}_2=x(t)={1\over 2}gt^2+v_0t+20=\text{distance}_1+20.$$ Thus, we see that shifting the launch point shifts the landing point by the corresponding amount.