Question

An object is thrown down with an initial velocity of 3.2 meters per second from a 52-meter tower. Assuming the position of the object above the ground is modeled by y(t) = -4.9t^2 - 3.2t + 52, find the velocity when the object hits the ground.

-32.085 m/s
-17.642 m/s
-2.947 m/s
2.947 m/s

Answer 1

Hey Bree :)
Is this a Calculus-related problem?
Recall that the derivative function of `position` gives us `velocity`, ya?

Answer 2

Hello! :)
Yes it is a calculus problem, I forget if I should find the first or second derivative? :o

Answer 3

Let's start by finding the time t, that corresponds to y=0.
y=0 is a height of zero, it's when the object hits the ground.\[\large\rm y(t) = -4.9t^2 - 3.2t + 52\]\[\large\rm \quad 0 ~= -4.9t^2 - 3.2t + 52\]

Answer 4

Or derivative first, which ever you prefer.
We won't need to worry about second derivative for this problem.

Answer 5

\[\large\rm \quad 0 ~= -4.9t^2 - 3.2t + 52\]Do you understand how to solve for t here?
It likely won't factor nicely,
so you'll have to throw it into your `Quadratic Formula`.

Answer 6

So the x coordinates would be -3.6 and 2.95? :)

Answer 7

Mmm k good!
We'll call it the t-coordinate since we're dealing with time.
And on that note, we're not concerned with anything before time zero,
we're not looking back into the past, no time machine.

So we'll ignore the t=-3.6 result that we got.

Answer 8

Good, so we found the `time` that corresponds to the object hitting the ground.

Answer 9

We would like the evaluate the velocity function `at this time`.
That will tell us how fast it was going when it hit the ground.

Answer 10

\[\large\rm y'(t)=v(t)=?\]So take your derivative :)
Power rule and such.

Answer 11

-24.01t - 3.2? :)

Answer 12

Hmm that first coefficient looks kind of large

Answer 13

-9.8 my bad :o

Answer 14

\[\large\rm y'(t)=-9.8t-3.2\]\[\large\rm v(t)=-9.8t-3.2\]Ok great.

Answer 15

\[\large\rm v(2.95)=?\]Just one final step then,
evaluating the velocity function at the specific time when the object hits the ground.

Answer 16

-32.11?? :)

Answer 17

Nay, set y(t) = 0 and solve for t
0 = 4.9*t^2 + 3.2*t - 52 <--- I multiplied by -1

Quadratic equation time
t = 2.95 s

Now just use basic kinematics
vf = vi - g*t

with vi = -3.2 m/s
vf = -3.2 - 9.81*2.95 = -32.1 m/s

The negative implies it is moving down

Answer 18

Ohhhh okay, that makes sense, thank you!! :)

Answer 19

np

Answer 20

-32.11 for your velocity?
yay good job \c:/

Answer 21

Thanks so much both of you! :D