A cannon is installed on a tower above an extensive, standard area. The barrel associated with the cannon is 180m over the surface the following. A cannonball is fired horizontally with a short rate of 600 m/s. Presuming environment opposition may be ignored, approx just how long will the cannonball take trip before it strikes the bottom?

1.Equation: Y=y(initial)+v(initial)t+1/2at^2

In which y= y element of equation, vvelocity, t=time, and a=acceleration of gravity

0=180+(0)t+1/2(-9.81)t^2

-180=-4.905t^2

t^2=36.69724771

t=6.06 moments

d) 6s

2. d=(v)(t)

In which d is length, v is velocity, and t is time

d=(600)(6.06)

d is about corresponding to 3600

c) 3,600