Wolfram Alpha:

```Kinematics
----------
Only spatial and time related variables.  Constant acceleration.

SUVAT:

a = (v - u)/t  => v = u +at

-
v = (u + v)/2  ... average velocity

s = (u + v)t/2

= (u + u + at)t/2

= ut + (1/2)at2

t = (v - u)/a

s = {(u + v)/2}{(v - u)/a} => 2as = -u2 + v2

=> v2 = u2 + 2as

Calculus:

v = ∫adt  ... area under a versus t curve

= u + at

s = ∫vdt  ... area under v versus t curve

= ∫( u + at)dt

= ut + (1/2)at2

s = (1/2)t(u + v)

2s = t(u + v)

t = (v - u)/a => 2s = {(v - u)/a}{(u + v)}

=> 2as = v2 - u2

=> v2 = u2 + 2as

In summary:

v = u + at
s = (1/2)(u + v)t
v2 = u2 + 2as
s = ut + (1/2)at2

where u = initial velocity, v = final velocity, s = distance, a = acceleration

Trajectories:

.    .
.      ^        .
.          |            .
u .            h             ^
/ θ            |              s
-------------R-v----------- -v---

Ex:  A ball is kicked from level ground at an angle 60 deg with initial
velocity of 10 m/s.

Range:

ux = 10cos60 => 5 m/s
uy = 10sin60 => 8.660 m/s

s = ut - (1/2)gt2  gravity is negative in equation
0 = 8.660t - (1/2)*9.8*t2  h = 0 same level
=> t = 17.32/9.8 = 1.767 s

R = uxt
R = 5*1.767 = 8.837 m

Peak height:

vy2 = uy2 - 2gs
h = u2sin2θ/2g    vy = 0
h = 8.6602/2*9.8 = 0.75/19.6 = 3.82 m

Time to peak:

vy = uy - gt
t = usinθ/g    vy = 0
t = 8.660/9.8 = 0.88 s

Alternatively, the vertex of the parabola with the equation
s = 8.660t - (1/2)*9.8*t2 can be found from t = -b/2a
(axis of symmetry)

t = -b/2a = 8.66/9.8 = 0.88 s
```