Question 844186
A cannon ball is fired almost vertically upwards from ground level.
 The cannon ball has height given by the relationship H = 12t – t^2 metres, where t is the time in seconds after firing.
Consider the following:
(i) If we sketch a graph of the height H against the time t after firing what shape will result?
• Draw an accurate sketch of this graph and label the graph in the context of this scenario.
I suggest you plot these points, replace x, find y
 x | y
-------
 0 | 0
 2 | 20
 4 | 32
 6 | 36
 8 | 32
 9 | 27
12 | 0 (ground level)
Look something like this
{{{ graph( 300, 200, -4, 15, -10, 100, -x^2+12x) }}} 
:
(ii) How long would it take for the cannonball to reach its maximum height?
• What would be the maximum height reached?
You can see it reaches max height when t = 6 sec (halfway between 0 and 12
• Re-write the relationship H in a different form using the values you obtained for the maximum height?
In the equation h = 12t - t^2, replace t with 6
h = 12(6) - 6^2
h = 72 - 36
h = 36m (as is shown on the graph)
:
(iii) The maximum height reached by the cannonball is doubled.
 Given that the cannonball will still land after the same flight time,
 write down a relationship for H, which would represent a doubling of the height.
Using the form y = ax^2 + bx + c, we have to find new values for a & b, (no c)
For the coordinates:
 x = 6, y = 72 (twice the height)
and
x = 12, y = 0 (same flight time)
write an equation for each pair replace x and y
36a + 6b = 72
and
144a + 12b = 0
multiply the 1st equation by 2, subtract from the 2nd equation
144a + 12b = 0
72a + 12b = 144
-----------------subtraction eliminates b, find a
72a = -144
a = -144/72
a = -2
Find b using the 1st original equation
36(-2) + 6b = 72
-72 + 6b = 72
6b = 72 + 72
6b = 144
 b = 144/6
 b = 24

Graph the new equation h = 24t - 2t^2
{{{ graph( 300, 200, -4, 15, -10, 100, -2x^2+24x) }}}
NOte that it reaches 72m in 6 sec now 
:
(iv) Investigate further variations in the height reached. Place this data in a table and reflect on your results.
Note that if you want to double the height and retain the same time
double the coefficients of t and t^2, for example to double the height again to
144m
48t - 4t^2
{{{ graph( 300, 200, -6, 15, -50, 150, -4x^2+48x) }}}
You can make the table here using this example
: 
(v) Investigate what would happen if the original height is maintained but the flight time is varied.
Using the form ax^2 + bx = y, find the a and b of the new equation
x=10, y=36; max occurs in 10 sec, but is still 36
x=20, y=0; hits the ground in 20 sec
100a + 10b = 36
400a + 20b = 0
Mult the 1st equation by 2, subtract from the 2nd
400a + 20b = 0
200a + 20b = 72
----------------subtraction eliminates b, find a
200a = -72
a = -72/200
a = -.36
Find b using the 1st original equation
100(-.36) + 10b = 36
-36 + 10b = 36
10b = 72
b = 7.2
the new equation: h = 7.2t - .36t^2, this graph looks like 
{{{ graph( 300, 200, -10, 24, -10, 50, -.36x^2+7.2x) }}}
max occurs at 10 sec but is still 36 m