Jele F.

asked • 01/02/17

The function represents the height y (in feet) of the paintbrush t seconds after it is dropped. After how many seconds does the paint brush land on the ground?

While standing on a ladder, you drop a paintbrush. The function represents the height y (in feet) of the paintbrush t seconds after it is dropped. After how many seconds does the paint brush land on the ground?
 
 
 
Given an equation: y=25-15t2 then an arrow down. Probably this function was how high the paintbrush was before it started falling?

Arturo O.

Jele,
 
Regarding your statement about the given equation that "Probably this function was how high the paintbrush was before it started falling?", we can say the following:
 
The given function
 
y(t) = 25 - 15t2
 
is a case of constant acceleration motion, described by
 
y(t) = y0 + v0t + (1/2)at2
 
where
 
y0 = 25 feet (i.e. initial height above ground, where height of ground is y = 0)
v0 = initial vertical speed = 0 ft/s (i.e. dropped from rest)
a = constant acceleration = -30 ft/s2
 
This equation gives the distance fallen from an initial height of 25 feet, starting from rest (initial speed = 0 ft/s), and accelerating downward at a rate of 30 ft/s2.  It may be more realistic to use
 
a = -(acceleration of gravity) = -g = -32 ft/s2
 
Then the function would be 
 
y(t) = 25 - 16t2
 
Andrew gave you a correct solution to the problem as it was stated in the description.  Just test it by plugging his answer for t into the given y(t) and see that it gives a height of 0 feet.
Report

01/02/17

1 Expert Answer

By:

Andrew M. answered • 01/02/17

Tutor
New to Wyzant

Mathematics - Algebra a Specialty / F.I.T. Grad - B.S. w/Honors

See tutors like this
See tutors like this
h(t) = y = -15t2 + 25
 
the height at ground level is zero.
Set the equation = 0 and solve for t.
 
-15t2 + 25 = 0
-15t2 = -25
t2 = -25/-15
t2 = 5/3
 
t = ±√(5/3) 
Note, we can discard the negative since it
will be a positive amount of time.
t = 1.291 seconds
Upvote • 1 Downvote
More
Report

Still looking for help? Get the right answer, fast.

Ask a question for free

Get a free answer to a quick problem.
Most questions answered within 4 hours.

OR

Find an Online Tutor Now

Choose an expert and meet online. No packages or subscriptions, pay only for the time you need.