Can I find acceleration due to gravity of a 2D projectile given only the peak of the projectile, the point of impact, initial angle, and initial velocity?

Specifically what is missing is the initial firing position. The firing position and landing position may be at different elevations.

Ignoring wind, air resistance, etc.

both $x$ and $y$ values for everything given for example:

Peak $(x,y)=(1000,500)$

Landing point $(x,y)=(2000,0)$

Initial Velocity = $100$

Launch Angle = $45$

  • $\begingroup$ Do we know the $x$ and $y$ values of the peak, or just $x$, or just $y$? What origin is/are it/they referenced to? Same question for the starting point and ending point. The answers to those questions make all the difference in the world! $\endgroup$
    – garyp
    Apr 6 '16 at 23:03
  • $\begingroup$ both x and y values for everything given for example: Peak X,Y = 1000, 500 Landing point X,Y = 2000, 0 Initial Velocity = 100 Launch Angle = 45 $\endgroup$
    – Luke Allen
    Apr 6 '16 at 23:22

If we assume that $g$ (acceleration due to gravity) is constant and parallel to our $y$ axis (but in the opposite direction), then we have: (Initial velocity=$v_0$, Launch angle=$\theta$, Initial position=$(x_0,y_0)$) $$a_y=-g\qquad (1)$$ $$\Longrightarrow \; v_y=-gt+v_0\sin \theta\qquad (2)$$ $$\Longrightarrow \; y=-\frac 12 gt^2+v_0 \sin \theta\; t+y_0\qquad(3)$$ $$a_x=0\qquad (4)$$ $$\Longrightarrow \; v_x=v_0 \cos \theta\qquad (5)$$ $$\Longrightarrow \; x=v_0 \cos \theta\; t+x_0 \qquad(6)$$ Finding $t$ from $(6)$ and replacing it in $(3)$ obtains: $$y=-\frac 12 g\frac{(x-x_0)^2}{(v_0\cos \theta)^2}+(x-x_0)\tan \theta+y_0\qquad (7)$$ So, we can find the $g$ if we have $x_0$, $y_0$, $v_0$, $\theta$ and an arbitrarily point position (like peak point) as $x$ and $y$.

But, as you have said in the problem description we don’t know the initial position of the projectile. What we know is the positions of two points of the projectile path that one of them is the peak point. We know that at peak point $v_y=0$.

Finding $t$ from $(6)$ and replacing it in $(2)$ obtains: $$v_y=-g\frac{x-x_0}{v_0\cos \theta}+v_0 \sin \theta\qquad (8)$$ Finally, we have three equations (two equations from $(7)$ and one equation from $(8)$) and three unknowns ($x_0$, $y_0$ and $g$) that we can determine all of them.

  • $\begingroup$ Thank you for your comment on my answer which explains why you had to work so much harder to produce an answer. So it has to be a +1 for you. $\endgroup$
    – Farcher
    May 17 '16 at 15:50

using the x and y components, find out the final velocity vector. The launch angle is 45 deg. At this angle, the distance traversed by the projectile is maximum. With this in mind, try to solve it.


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