Question

The equation h(t)=−16t2+19t+110 gives the height of a rock, in feet, t seconds after it is thrown from a cliff.

Answer 1

We know that
$h(t)=( \frac{1}{2} )a t^{2} +v0t+h0$

 Where: 
 a: is the acceleration
 vo: is the initial velocity
 h0: is the initial height

in this problem
a=-32 ft/sec²
v0= 19 ft/sec
h0=110 ft
 Substituting values we have:
$h(t)=-16 t^{2} +19t+110$
  The initial velocity is : vo = 19 feet / s


 The answer is
 The initial velocity is  vo = 19 feet / s

Answer 2

For this case we have an equation of the form:
 $h (t) = (a / 2) * t ^ 2 + vo * t + h0$
 Where,
 a: acceleration
 vo: initial speed
 h0: initial height.
 In this case we have the following equation:
 $h (t) = - 16t ^ 2 + 19t + 110$
 Therefore, the initial velocity is:
 $vo = 19 ft / s$
 Answer:
 The initial velocity when the rock is thrown is:
 vo = 19 ft / s