You throw a ball vertically upwards with an initial speed `v (in km per hour)`. The height `h` of the ball at each time `t`
is given by `h = v*t - 0.5*g*t*t` where `g` is Earth's gravity `(g ~ 9.81 m/s**2)`. A device is recording at every **tenth
of second** the height of the ball.
For example with `v = 15 km/h` the device gets something of the following form:
`(0, 0.0), (1, 0.367...), (2, 0.637...), (3, 0.808...), (4, 0.881..) ...`
where the first number is the time in tenth of second and the second number the height in meter.
# Task
Write a function `max_ball` with parameter `v (in km per hour)` that returns the `time in tenth of second`
of the maximum height recorded by the device.
# Examples:
`max_ball(15) should return 4`

`max_ball(25) should return 7`
# Notes
- Remember to convert the velocity from km/h to m/s or from m/s in km/h when necessary.
- The maximum height recorded by the device is not necessarily the maximum height reached by the ball.

Source: codewars | Difficulty: EASY
Tags: Fundamentals

Starter code: def max_ball(v0):
	

Solutions:
def max_ball(v0):
	return round(10 * v0 / 9.81 / 3.6)


def max_ball(v):
	return round(v / 3.5316)


def max_ball(v0, g=9.81):
	v = v0 / 3.6
	prev = 0
	for i in range(1000):
		t = i * 0.1
		h = v * t - 0.5 * g * t * t
		if h < prev:
			return i - 1
		prev = h


def max_ball(v0):
	g = 9.81

	def h(u):
		return v * u - 0.5 * g * u * u
	v = v0 * 1000 / 3600.0
	t = v / g
	tm = int(t * 10)
	h1 = h(tm / 10.0)
	h2 = h((tm + 1) / 10.0)
	mx = max(h1, h2)
	if mx == h(tm / 10.0):
		return tm
	else:
		return tm + 1


max_ball = lambda v0: round(v0 * 1000 / 360 / 9.81)


def max_ball(v0):
	v = v0 * (1000 / 3600)
	g = 9.81
	t = 0
	record = []
	while v * t - 0.5 * g * t * t >= 0:
		h = v * t - 0.5 * g * t * t
		record.append(h)
		t += 0.1
	return record.index(max(record))


from itertools import count

def max_ball(v0):
	g = 9.81
	c = count(0, 0.1)
	hh = -0.1
	while True:
		t = next(c)
		h = v0 * 1000 / 3600 * t - 0.5 * g * t * t
		if h - hh > 0:
			hh = h
		else:
			return int(round(t * 10, 2)) - 1