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Serga [27]
3 years ago
10

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

Mathematics
2 answers:
Vlada [557]3 years ago
6 0
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
Butoxors [25]3 years ago
3 0
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
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