a) at it's maximum height dy/dx = 0 so -24x+144 = 0 and x = 6
b) when the object hits the ground y = 0 so -12X^2 + 144x + 6 = 0 - solve using quadratic formula
c) put y=225 into the equation and solve the quadratic. The difference between the two 'x' values is the time it is above 225
d) set x =...
