Question

A object is thrown straight upwards with a speed of 19.6 m/sec. Three seconds later, its speed is:
Select one:
a. zero.
b. 9.8 m/sec downward.
c. 49 m/sec upward.
d. 29.4 m/sec downward.

Answer 1

Okay, do you know the downward acceleration due to gravity?

Answer 2

Nope it doesnt say

Answer 3

That is because you're supposed to memorize it. It is always \(9.80665 m/s^2\).

Answer 4

You can round it to \(9.81m/s^2\) for this problem.

Answer 5

So you want to change from \(\large \frac{m}{s^2}\) to \(\large \frac{m}{s}\)

Answer 6

That's a hint that you usually want to multiply by \(s\).

Answer 7

oh ok

Answer 8

If you accelerate at a constant rate for a period of time, the change in velocity is acceleration times time.

Answer 9

So first, what is the change in velocity?

Answer 10

um... Vf-Vi

Answer 11

Yes, that is true. But we want to calculate it by using acceleration and time, since that is what we know.

Answer 12

so vf=Vi+acceleration(time)
\[Vf = Vi+ a (t)\]

Answer 13

Yeah. We are given time and initial velocity (and we keep acceleration to memory).

Answer 14

So we should be able to figure this out.

Answer 15

Yay :D

Answer 16

the answer i got wasnt in any of the choices

Answer 17

Oh, well one thing to note is that gravity is downwards.

Answer 18

so its -9.8 m/s^2

Answer 19

So it will be in the opposite direction of the initial velocity.

Answer 20

That is one way of doing it.

Answer 21

? how would it be opposite?

Answer 22

You can chose upward to be positive and downward to be negative. You can also choose upward to be negative and downward to be positive. As long as you are consistent.

Answer 23

The initial velocity is upward, gravity is downward. They are in the opposite direction.

Answer 24

yeah but either way i got 19.6m/s+-88.2m

Answer 25

\[Vf = 19.6\frac{ m }{ s } +( -9.8 \frac{ m }{ s^2 } ) (3^2) \]

Answer 26

That 3 shouldn't be squared!

Answer 27

Omg i just noticed THANK YOU