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?

Guest Feb 5, 2020

#1**+1 **

maybe just set this function equal to 6 and solve for t, the result will be two cases.

t1 nearly equal 0.57s

t2 nearly equal 2.29s

hope it can help :)

noobfromvn26 Feb 5, 2020

#2**+2 **

6 = -4/9t^2 + 14t - 0.4 rearrange as

4.9t^2 - 14t + 6.4 = 0 multiply through by 10

49t^2 - 140t + 64 = 0 factor

(7t - 4) ( 7t - 16) = 0

Set each factor to 0 and solve for t and we have that

t = 4/7 sec and t = 16/7 sec

So.....the time it is above 6 m is

(16/7) - (4/7) =

12/ 7 sec

CPhill Feb 5, 2020