SOLUTION: A ball is tossed from a window that is 10 feet off the ground. The height, h (in feet), of the ball at any time, t (in seconds), is given by the equation h = - 16 t^2 + 12x +

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Question 635476: A ball is tossed from a window that is 10 feet off the ground. The height, h (in feet), of the ball at any time, t (in seconds), is given by the equation h = - 16 t^2 + 12x + 10. What is the maximum height (in feet) that the ball will reach?
Found 2 solutions by lwsshak3, ewatrrr:
Answer by lwsshak3(11628)   (Show Source): You can put this solution on YOUR website!
A ball is tossed from a window that is 10 feet off the ground. The height, h (in feet), of the ball at any time, t (in seconds), is given by the equation h = - 16 t^2 + 12x + 10. What is the maximum height (in feet) that the ball will reach?
**
h = - 16 t^2 + 12x + 10
This is an equation of a parabola that opens downwards (function has a maximum)
Its standard form: y=(x-h)^2+k, (h,k)=(x,y) coordinates of vertex, k=maximum value
complete the square
h=-16(t^2-(12/16)t)+10
=-16(t^2-(3/4)t)+10
=-16(t^2-(3/4)t+9/64)+9/4+10
=-16(t-(3/8)^2+9/4+40/4
h=-16(t-(3/8)^2+49/4
maximum height that the ball will reach=12.25 ft
Answer by ewatrrr(24785)   (Show Source): You can put this solution on YOUR website!
 

Hi,

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

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

h = - 16 t^2 + 12x + 10 

  

h = -16(x - 3/8) + 16(9/64) + 10 

h = -16(x - 3/8) + 9/4 + 40/4    \ parabola opening Downward  a = -16 < 0

h = -16(x - 3/8) + 49/4  Vmax(3/8, 12.25)

12.25 ft the maximum height that the ball will reach





 

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