Question

the trajectory of a projectile is given by the equation y=-x squared+200x.find the value of x at which it hits the ground

Answer 1

Solve your equation for when y=0:\[0=-x ^{2}+200x\]You'll get two possible answers but only one is correct.

Answer 2

Of the two answers you will see that one of them is x=0
(just put 0 into the above equation to see that this is true)

This is the 'start position- when the projectile is at ground level prior to being launched.

The second solution is when it returns to ground level after its trajectory.

In most cases - if you have a quadratic then BOTH solutions have a sensible meaning

Answer 3

?ok where do i go from there