Question

If an arrow is shot upward on the moon with velocity of 47 m/s, its height (in meters) after t seconds is given by
h(t)=47t−0.83t2.


(a) Find the velocity of the arrow after 8 seconds.
Velocity =
(b) Find the velocity of the arrow after a seconds.

Velocity =
(c) When will the arrow hit the moon?

Time =
(d) With what velocity will the arrow hit the moon?

Velocity =

Answer 1

\[\large h(t)=47t-0.83t^2\]\[\large v(t)=h'(t)\]

Answer 2

so find the derivative

Answer 3

ya.

Answer 4

After you find the derivative,
You'll notice that if you plug t=0 in, it matches the information they gave us at the start.

\[\large v(0)=47\]The initial velocity is 47m/s as they told us. So that let's us know we're on the right track.

Answer 5

ok so i got that but as far as the next too?

Answer 6

b) and c)?

Answer 7

c and d

Answer 8

When will the arrow hit the moon. So the question is asking, when will the `height` be 0.

Well... we know it's 0 at the start right? Because the arrow is launched from the ground. But when will it hit the ground again?

It will hit the ground when \(\large h(t)=0\).
So we'll set it equal to zero, and solve for t.
This is equivalent to finding "roots" of a polynomial if you can remember that :)

\[\large 0=47t-0.83t^2\]

Answer 9

56.6265

Answer 10

so then how do we get d

Answer 11

So you determined that the arrow will hit the ground at time \(\large t=56.6265s\).

So we'll plug that time into the velocity function to determine how FAST the arrow is going when it hits the ground at this time.

\(\large v(56.6265)=?\)

Answer 12

-46.99999 thank you