+0  
 
+1
98
1
avatar+500 

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?

 Apr 13, 2021

Best Answer 

 #1
avatar+33727 
+2

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

\(t = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

 Apr 13, 2021
 #1
avatar+33727 
+2
Best Answer

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

\(t = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

ElectricPavlov Apr 13, 2021

15 Online Users

avatar