Answer:
A ball is thrown at an initial height of 5 feet with an initial upward velocity at 29 ft/s. lets assume that  balls height h (in feet) after t seconds is give by: 
<u>h= 5 + 29t -16t^2</u>
Explanation:
h= 5 + 29t -16t^2
  
 a time when the ball's height will be 17 ft
  
17 = 5 + 29t -16t2
  
0 = -17 + 5 + 29t -16t2
  
0 = -12 + 29t - 16t2
  
Using the quadratic equation:
  
t = (-29±√(292-(4*(-16)*(-12))))÷2(-16)
  
  = (-29±√(841 - 768))÷(-32)
  
  = (-29±√(73))÷(-32)
  
  = (-29 + 8.544)÷(-32)   or   (-29 - 8.544)÷(-32)
  
  = (-20.456)÷(-32)         or   -37.544÷(-32)
  
  =  0.64                         or   1.17
  
So, the ball is at a height of 17 ft twice: once on the way up after 0.64 seconds and once on the way back down after 1.17 seconds.