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

2 answers:

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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Answer:

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Step-by-step explanation:

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We can do this as follows:

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