Question is in the description

Question

A boy standing on the top of a building throws a water balloon up vertically. The height,h(infeet),ofthewaterballoonisgivenbytheequationh(t)=−16t2 +64t+192,where t is the time in seconds after he threw the water balloon. What is the value of t when the balloon hits the ground? Explain

Answer 1

In order to find the value of t when the balloon hits the ground we need to set h(t) =0 since it represents the height above the ground so when it hits the ground it has a height of zero.

 

 

-16t^2 + 64t + 192 = 0

 

multiply both sides by minus 1

 

16t^2 - 64t - 192 = 0

 

divide by 16 

 

t^2 - 4t - 12 =0

 

Now solve the quadratic by factoring

 

(t - 6) ( t + 2) = 0

 

our two values of t are -2 and 6 seconds however time cannot be negative thus our answer is 

 

t = 6 seconds.

 

Answer 2

The balloon hits the ground when h(t) = 0.

 

0 = -16t² + 64t + 192              factor out -16

0 = -16(t² - 4t   ) + 192           complete the square

0 = -16(t² - 4t + 2² ) + 192 + 64

0 = -16(t - 2)² + 256               subtract 256 from both sides

-256 = -16(t - 2)²                    divide both sides by -16

16 = (t - 2)²                           take square root of both sides

±4 = t - 2                               add 2 to both sides

2 ± 4 = t

 

The negative value is discarded as meaningless.

 

The balloon hits the ground at six seconds.