A cannon ball is shot horizontally off a 37.0 m cliff and lands a distance of 18.5 m from the base of the cliff. Whall was the initial horizontal velocity of the cannon ball?

vₓ = 6.73 m/s

Explanation:

Assuming no other external influences than gravity, in the horizontal direction (which we make to coincide with the x- axis) , speed is constant, so, applying the definition of average velocity, we can write the following equation:

       

Now, in the vertical direction (coincident with the y- axis) , as both movements are independent each other, initial velocity is zero, so we can write the following equation for the vertical displacement:

       

where Δh = -37.0 m , g = -9.8 m/s2Solving (2) for t, we get:

           

Taking t₀ = 0, ⇒ Δt = tReplacing (3) in (1), we get:

       

As the horizontal velocity is constant, the initial horizontal velocity is just the average one, i.e., 6.73 m/s.