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.

What is the initial velocity when the rock is thrown?

Enter your answer in the box.


= ft/s

Answer 1

we know that

`h(t)=( (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