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

What is the initial velocity when the rock is thrown?

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= ft/s

Respuesta :

carlosego
For this case we have an equation of the form:
 [tex]h (t) = (a / 2) * t ^ 2 + vo * t + h0 [/tex]
 Where,
 a: acceleration
 vo: initial speed
 h0: initial height.
 In this case we have the following equation:
 [tex]h (t) = - 16t ^ 2 + 19t + 110 [/tex]
 Therefore, the initial velocity is:
 [tex]vo = 19 ft / s [/tex]
 Answer:
 The initial velocity when the rock is thrown is:
 vo = 19 ft / s
calculista
we know that
[tex]h(t)=( \frac{1}{2} )a t^{2} +v0t+h0[/tex]

 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:
[tex]h(t)=-16 t^{2} +19t+110[/tex]
  The initial velocity is : vo = 19 feet / s


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

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