## Finding the velocity with out time (s)

1. The problem statement, all variables and given/known data
A launcher is on top of a table and fire the projectile horizontally. The table/launcher height is 1.1 m and the projectile landed 3.2 m away from the table.
h=1.1 m
R=3.2 m

Find the velocity of the projectile as it leaves the launcher.

2. Relevant equations
tanθ=y/x
Δx = V(i)t + 1/2at^2

3. The attempt at a solution

θ = tan^-1(1.1/3.2) θ= 18.97°
After this I am lost. I assume I need the angle but I'm not sure for what.

I think I may need to solve for both x and y. My x and y values are (0,1.1) (3.2,0)

x = V(i)t ?
y = h -1/2gt^2 ?

Since I don't have t I don't what other equations I can use.

Please let me know if I am on the right track!
Thank you.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

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 On the x-y coordinates system, make a vertical and a horizontal lines. The vertical line starts from zero and moves according to velocity. The horizontal line moves from zero according to velocity and acceleration too. The intersection of these lines determine the position of the projectile. Remember it is only one projectile.
 From the equation y(t)=0 you obtain the time needed by the projectile to land, call this time for example $t_\ell$. Then because $x(t_\ell)=R$ then from the equation $R=V(i)t_\ell$ you can find V(i).

## Finding the velocity with out time (s)

Hint: Read the question and look what it given.

 And if the projectile was fired horizontally, the initial velocity in the vertical component is 0. So, look at all your kinematic equations, solve for t and then solve for initial velocity in horizontal component.