Quadratic word problem

Question

a person standing at the edge of a 32 meter building throws a baseball vertically upward. The quadratic function h(t)= —16t^2 + 64t + 32 models the balls height above the ground , h(t) in meters , t seconds after it was thrown. 

 

After how many seconds does the ball reach its maximum height and what is the maximum height ? After how many seconds does the ball hit the ground?

Answer 1

This equation can be rewritten as an upside down parabola.

 

 y-h=4p(x-k)²

 

vertex form of the parabola.

 

y=-16t2 + 64t - 32

 

 y - 32=-16(t^2- 4t)

 

Complete the square

 y - 32 - 64=-16(t^2 -4t + 4)

Had to put in term of -64 or -16*4 on left hand side of equation.

 

y - 96=-16(t - 2)^2

 

When t=2, y=96.  After 2 seconds, maximum height will be 96 feet.

 

When ball hits ground, y=-32.

 

Put -32 into equation.

 

-32 - 96=-16(t -2)^2

-128=-16(t - 2)^2

4=(t - 2)^2, we divided both sides by -16.

 

-2=t - 2, 2=t - 2

 

t=0, 4.  Choose t=4 because at t=0, ball is at top of the building.