you throw a snowball horizontally. it hits a target 2 meters away. you notice that it hit .9 m below were u aimed. with what horizontal velocoty did you throw the ball

The target is 2 meters ahead of you.

If you throw the snowball with a velocity S, then the time that the snowball will need to hit the target is:

2m/S = T

Now let's analyze the vertical problem.

The only force acting on the snowball is the gravitational force, then the acceleration will be -g = -9.8m/s^2.

a(t) = -9.8m/s^2

For the vertical velocity, we need to integrate over time and get:

v(t) = (-9.8m/s^2)*t + v0

Where v0 is the initial vertical velocity, in this case is zero because the ball is thrown horizontally.

v(t) = (-9.8m/s^2)*t

For the vertical position, we integrate again over time.

p(t) = (-4.9m/s^2)*t^2 + p0.

Now, we know that when the ball hit the target, it hits 9 meters below the desired point (where the desired point is p0), so p(2m/S) = p0 - 9m.

Then we have:

(-4.9m/s^2)*(2m/S)^2 = -9m.

We need to solve this for S.

(-4.9m/s^2)*4m^2/-9m = S^2

2.178 m^2/s^2 = S^2

√(2.178 m^2/s^2) = S = 1.48 m/s

The horizontal velocity is 1.48 m/s