Solving assymetrical projectile motion problems can be either

CROSSFIRE: ASYMMETRICAL PROJECTILE MOTION
Solving asymmetrical projectile motion problems can be either relatively straight forward, or require some very messy math to find the solution. They all, however, start with the same basic conditions and boil down to one formula.
The two possibilities are you are at a certain elevation h and wish to launch a projectile at angle θ and velocity v to cover range d (fig. 1) , or you are at ground level at range d and wish to launch a projectile at angle θ and velocity v to reach elevation h (fig. 2)
For both cases, the height h reached after time t is given by
()02
2
1sin h t v at
h ++=
θ
Also, the range d can be found by
()t v d θcos =
Which can be rearranged and substituted into the first equation for t
or
()02
21tan cos h d v d a h ++⎪⎭
⎫
⎝⎛=θθ
If you take a =–9.8m/s 2=–g , then the general formula for finding h 0 or h (assuming the other is zero) is
()2
210cos tan ⎪⎭
⎫
⎝⎛-=-θθv d g d h h
d Fig. 2
Fig. 1
()0
221cos sin cos h v d v v d a h +⎪⎭
⎫
⎝⎛+⎪⎭⎫ ⎝⎛=θθθ
If you are given the initial velocity and angle, solving for height or range is simple. But what if you know the elevation you are starting from or the elevation above you that you wish to reach, but you only know the initial velocity , not the angle?
This cannot be so easily solved, because the angle θ is part of two different trig functions. To proceed, you must
either choose an algebraic approach or a trigonometric approach : both involve quadratic solutions, because two angles will satisfy the conditions, as you have seen with the range formula. Many thanks to Andrew Rothfuss and Haoyang Liu for their assistance in creating these notes.
Algebraic approach
First, call h-h 0=y, d=x, vsinθ=v y and call vcosθ=v x; ; it is also known that 2
22
x y v v v +=
Then the previous general-purpose solution becomes
2
21⎪⎪⎭
⎫ ⎝⎛-⎪⎪⎭⎫
⎝⎛=x
x y
v x g v v x y solving for v x and then substituting
x y v v v =-2
2
is very tedious;
()()()()()()()()()()
(
)()0
10210
210
222
212
2
2
222
212
2212
222
212
21222
22
212122
2
2121
2
2
22
21
2
22
122
2
12
2
1
=+
-+⎥⎥⎦
⎤⎢⎢⎣⎡+⎪⎭⎫ ⎝⎛=+⎥⎦⎤⎢⎣⎡-⎪⎭⎫ ⎝⎛+⎥⎥⎦⎤⎢⎢⎣⎡+⎪⎭⎫ ⎝⎛=⎥⎦⎤⎢⎣⎡+-⎪⎭⎫ ⎝⎛+⎥⎥⎦⎤⎢⎢⎣⎡+⎪⎭⎫ ⎝⎛=+-⎥⎦
⎤⎢⎣⎡+⎪⎭
⎫
⎝⎛+⎪⎭⎫ ⎝⎛⎥
⎦
⎤⎢⎣⎡+⎪⎭⎫ ⎝⎛+⎥⎦⎤⎢⎣⎡⎪⎭⎫ ⎝⎛=-⎥⎦
⎤⎢⎣⎡+⎪⎭
⎫ ⎝⎛=-+⎪⎭
⎫ ⎝⎛=-+⎪⎭
⎫ ⎝⎛=-=gx v
v
yg v x y gx v v gx x y v x y v gx v gx x y v x y v v v gx gx x y v x y v gx gx x y v x y v v v gx v x y v v v gx
v x y v
v v gx
v x y v v gx xv yv x
x x x x x x x x x x x
x x x
x x x
x
x y
x
y x
Fig. 1 d
Fig. 2
it is much more productive to solve for v y and substitute
y x v v v =-2
2
x 2, where
solve for the angle θ=cos -1(v x /v)
Trigonometric approach
Start with the general-purpose formula
()()()()()
()()
()
()()0
tan tan tan tan 1
tan tan sec tan cos 1tan cos tan 02
21
22212
21
222102
2
2102
2
2102
2
2102
210=-+⎥⎥⎦
⎤⎢⎢⎣⎡⎪⎭⎫ ⎝⎛+-⎥⎥⎦⎤⎢⎢⎣⎡⎪⎭⎫ ⎝⎛⎥⎥
⎦
⎤⎢⎢⎣⎡⎪⎭⎫ ⎝⎛-⎥⎥⎦⎤⎢⎢⎣⎡⎪⎭⎫ ⎝⎛-=-+⎪⎭
⎫ ⎝⎛-=-⎪⎭
⎫ ⎝⎛-=-⎪
⎭⎫
⎝⎛⎪⎭⎫ ⎝⎛-=-⎪
⎭
⎫
⎝⎛-=-h h v d g d v d g v d g v d g d h h v d g d h h v d g d h h v d g d h h v d g d h h θθθθθθθ
θθθθθ
Use the quadratic equation to solve for tanθ, where
0-1。