View Full Version : Trajectory - quadratic equation

05-30-2006, 02:29 PM
The trajectory of an object moving through space is given by the quadratic equation:

d=-t squared + 6t

the value d represents the height off the ground of the projectile (in feet). The variable t represents the time ( in seconds)

How many seconds does it take the object to reach the highest point of its trajectory? What is the value of d at the highest point of trajectory?

How long does it take the object to reach the ground after it is launched.

05-30-2006, 03:33 PM
Someone should have told you two things:

1) Solve for t, using d = 0, to get time on the ground. t=0 should be one solution. The other solution is the answer to your last question.

2) The highest point is always right in the middle.

05-30-2006, 04:01 PM
5 feet in 2 seconds

d=t squared + 5t

05-30-2006, 04:03 PM
I have no idea what you mean, since you didn't show anything.

Start with the original equation.
Substitute d = 0 and solve for 't'.

No shortcuts.
Show your work.