The vertical movement of the projectile is described by:
y = H - gt² / 2
When the cannonball is on the ground, y= 0 so:
0 = H - gt² /2
Solving for t:
[tex]t = \sqrt{\dfrac{2H}{g}} = \sqrt{\dfrac{2(40\;m)}{9.8\;m/s^2}} = 2.8\;s[/tex]
The horizontally movement of the projectile is described by:
x = v₀t
Solving for v₀:
v₀ = x/t
v₀ = 175 m / 2.8 s
v₀ = 61.5 m/s
