i want to know im right.

my answer: x= 16t^2 + 9.8m/s

here is the question: An object, with velocity v0 m/s falls with constant acceleration 9.8m/s. how fast it goes after falling x m?

Nope, you are wrong.

Vf=vo+g*t

I have no idea how you got your answer.

bobpursley i did it wrong. thank you

To find the final velocity after the object falls a distance of x meters, we can use the equation of motion for an object undergoing constant acceleration.

The equation is: v^2 = v0^2 + 2as

Where:
v is the final velocity
v0 is the initial velocity (in this case, the object is starting from rest, so v0 = 0)
a is the acceleration (in this case, the constant acceleration due to gravity is 9.8 m/s^2)
s is the distance (in this case, x meters)

Plugging in the values, we have:
v^2 = 0^2 + 2 * 9.8 * x
v^2 = 0 + 19.6x
v^2 = 19.6x

To solve for v, we take the square root of both sides of the equation:
v = √(19.6x)

So, the correct answer is v = √(19.6x), not x = 16t^2 + 9.8m/s.