+501 The height (in meters) of a shot cannonball follows a trajectory given by $h(t) = -4.9t^2 + 14t - 0.4$ at time $t$ (in seconds). As an improper fraction, for how long is the cannonball above a height of $6$ meters?
+37146 put '6' in for height in the equation given ( which is a dome shaped parablola due to the negative coeeficient of t2)
6 = -4.9 t^2 + 14t - .4
0 = - 4.9t^2 + 14t - 6.4 now use Quadratic Formula to calculate the two 't values
....the ball is above 6 between thes two values , so calculate the time between these two values of t
a = -4.9 b = 14 c = -6.4
\(t = {-b \pm \sqrt{b^2-4ac} \over 2a}\)
+37146 put '6' in for height in the equation given ( which is a dome shaped parablola due to the negative coeeficient of t2)
6 = -4.9 t^2 + 14t - .4
0 = - 4.9t^2 + 14t - 6.4 now use Quadratic Formula to calculate the two 't values
....the ball is above 6 between thes two values , so calculate the time between these two values of t
a = -4.9 b = 14 c = -6.4
\(t = {-b \pm \sqrt{b^2-4ac} \over 2a}\)
