How are the endpoints of motion determined?
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Simply use the standard equation of motion
[math]s=ut+\tfrac12at^2[/math]
for each dimension: horizontal ([math]x[/math]) and vertical ([math]y[/math]).
Horizontally (ignoring air resistance) there is no acceleration so [math]a=0[/math] and
[math]x=t\cdot u\cos\theta[/math]
where [math]\theta[/math] is the launch angle.
Vertically acceleration is due to gravity so [math]a=-g[/math] and
[math]y=t\cdot u\sin\theta-\tfrac12g\cdot t^2[/math]
This provides the coordinates parameterised in terms of time ([math]t[/math]) and the initial speed ([math]u[/math]) assuming the launch point is the origin.
The other end of the trajectory also has [math]y=0[/math] so
[math]t=\frac{2u\sin\theta}{g}[/math]
[math]\Rightarrow x=\frac{u^2\cdot2\sin\theta\cos\theta}{g}=\frac{u^2\cdot\sin{2\theta}}{g}[/math]
[math]s=ut+\tfrac12at^2[/math]
for each dimension: horizontal ([math]x[/math]) and vertical ([math]y[/math]).
Horizontally (ignoring air resistance) there is no acceleration so [math]a=0[/math] and
[math]x=t\cdot u\cos\theta[/math]
where [math]\theta[/math] is the launch angle.
Vertically acceleration is due to gravity so [math]a=-g[/math] and
[math]y=t\cdot u\sin\theta-\tfrac12g\cdot t^2[/math]
This provides the coordinates parameterised in terms of time ([math]t[/math]) and the initial speed ([math]u[/math]) assuming the launch point is the origin.
The other end of the trajectory also has [math]y=0[/math] so
[math]t=\frac{2u\sin\theta}{g}[/math]
[math]\Rightarrow x=\frac{u^2\cdot2\sin\theta\cos\theta}{g}=\frac{u^2\cdot\sin{2\theta}}{g}[/math]
Answered by
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Simply use the standard equation of motion [math]s =ut+\tfrac12at^2[/math] for each dimension: horizontal ([math]x[/math]) and vertical ([math]y[/ math]).
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