Question

A cannon ball is shot horizontally off a 37.0 m cliff and lands a distance of 18.5 m

Answer 1

Answer:

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:

       $v_{x} = \frac{\Delta x}{\Delta t} (1)$

• 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:

       $\Delta h = \frac{1}{2} * g * t^{2} (2)$

• where Δh = -37.0 m , g = -9.8 m/s2
• Solving (2) for t, we get:

       $t = \sqrt{\frac{2*\Delta h}{g} } =\sqrt{\frac{2*37.0m}{9.8m/s2}} = 2.75 s (3)$    

• Taking t₀ = 0, ⇒ Δt = t
• Replacing (3) in (1), we get:

       $v_{x} = \frac{\Delta x}{\Delta t} = \frac{x}{t} = \frac{18.5m}{2.75s} = 6.73 m/s$

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