Question 3274
 h=-16 t^2 +vt+ s, s = 0, 
 
 1.a)  v = 40 ft/s
  When h = 24, solve the equation 24 = -16 t^2 + 40t or
   2t^2 - 5t + 3 = 0, 
  Factoring (2t -3)(t - 1) = 0, so t = 1 or 3/2
  Hence, when t = 1 or 1.5 sec, h = 24 ft
  b) The height of the ground is h = 0, solve
     -16 t^2 + 40t = 0 or 8t(2t -5) = 0.
    So, t = 5/2 or 0(means initial)
   Hence, when t = 2.5 sec, it hits the ground.
 
 2.a)  v = 64 ft/s
  When h = 48, solve the equation 48 = -16 t^2 + 64t or
   t^2 - 4t + 3 = 0, 
  Factoring (t -3)(t - 1) = 0, so t = 1 or 3
  Hence, when t = 1 or 3 sec, h = 48 ft
  b) The height of the ground is h = 0, solve
     -16 t^2 + 64t = 0 or 16t(t -4) = 0.
    So, t = 4 or 0(means initial)
   Hence, when t = 4 sec, it hits the ground.


 Kenny