Hi,

The height h (in meters) of a cannonball t seconds after it is fired from a cannon is described by the equation

 h(t)=-49t^2+60t.

 a. What is the height of the cannonball after 4 seconds?   h(4) = -544m

 b. How long will the cannonball be in the air? 1.22 sec

    0 = -49t^2 + 60t = t(-49t + 60)

    49t= 60

      t = 60/49 = 1.22 sec



Completing Square:  y = ax^2 + bx + c   ⇒ y = a(x -(-b/a2)^2 - a(-b/2a)^2 + c

   ,   V(60/98, 18.3673)

the vertex form of a Parabola opening up(a>0) or down(a<0), 

 where(h,k) is the vertex and the Line of symmetry is x = h

 c. How long does it take the cannonball to reach its maximum height? 60/98 sec

 d. What is the maximum height of the cannonball? 18.3673m



 

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