The vertical height is h = 180 ft
The acceleration due to gravity is g = 32.2 ft/s²
Let t = the time for the cannonball to hit the ground.
Because the initial vertical velocity is zero (the ball is dropped), the formula to use is
h = (1/2)gt²
That is
t = √[(2h)/g]
= √[(2*180)/32.2]
= 3.344 s
Answer:
The formula is [tex]t= \sqrt{ \frac{2h}{g} } [/tex], and the time is 3.344 s
