The height of an object 'x' seconds after it is launched is represented by...

B

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...the equation y=-12x^2+144x + 6? a) At what time, to the nearest second is the object at it's maximum height?
b) At what time, to the nearest second, will the object hit the ground?
c) for how long, to the nearest tenth of a second, is the object above 225 feet?
d) From what height was the objest launched?
 
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 = 0 and you get y(0) = 6
 
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 = 0 and you get y(0) = 6
 
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 = 0 and you get y(0) = 6
 
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