A cannon ball is fired straight up into the air at a speed of 80 ft/sec from a height of 2 ft. a)when will be the cannon ball be 98 feet in the air. b) when will the cannon ball hit the ground?

Answer 1

the equation for the ball's height at time t will be

h = 2 + 80t - 16t^2

So, it will be 98' up when t = 2

h=0 when t = 5.025 sec
(at t=5, it will be back to its original 2' height)