Asked by muthu
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?
Answers
Answered by
Steve
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)
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)
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