If a cannonball is shot directly upward with a velocity of 160 ft per second, its height above the ground after t seconds is given by s(t)equals160 t minus 16 t squared. Find the velocity and the acceleration after t seconds. What is the maximum height the cannonball reaches? When does it hit the ground? The velocity after t seconds is v(t)equals 160 minus 32 t ft/sec. The acceleration after t seconds is a(t)equals negative 32 ft divided by sec squared. The cannonball reaches a maximum height of 400 ft. The cannonball hits the ground after tequals 10 sec.