Question

can someone help me with 17-20 really dont uunderstand

Answer 1

@mathmath333

Answer 2

can you please help me. Really confused and teacher never explain on how to do it

Answer 3

i may be wrong, for \(17.\)

use \(s(t)=0\), \(t=4.5\)

Answer 4

i was thinking about doing that but im not sure how to finish

Answer 5

what is the value of \(v_0\)

Answer 6

yeah i really dont know teacher never explained it.

Answer 7

does this involves calculus.

Answer 8

yes

Answer 9

\[v(t) = s'(t)\]

Answer 10

\[v(t) = \frac{d}{dx} (s(t)=16t^2+v_0t+s_0)\]

Answer 11

yeah im really lost

Answer 12

I think you also need to apply your constant acceleration equations to find the final velocity. Particularly \(v_{f_y}^2 = v_{i_y}^2 +2g\Delta y\)

Answer 13

i never seen that equation before

Answer 14

HEY you can find the time first by setting the position function = 0 can you not?

Answer 15

i really dont know im sorry

Answer 16

At t=0 , h = 150 m
\[s(t) = 16t^2 +v_0t+150=0\]initial velocity is 0, because it is being dropped, not thrown. \[16t^2+150=0\]

Answer 17

for #26 i got the initial velocity was 150

Answer 18

#26?

Answer 19

#16

Answer 20

Oh. How can initial velocity be 150? can you explain?

Answer 21

what i did was make so=150 and t=0 -16xo^2 =vo(t) +0 t=150 so intial velocity =150 that what i think

Answer 22

sorry the site crashed on me

Answer 23

t is not the initial time,
its the time taken for object to change the position

so t can't be taken as 0 :)

Answer 24

Haha yeah, imagine if the ball were to magically drop from your hands and appear on the ground in t=0 seconds. That would be crazy.

Answer 25

\(-
16t^2+150=0\)

Answer 26

Why is it -16t\(^2\)? Is this because the position is moving downwards and we have labeled the positive position as up?

Answer 27

that what my tutor told me . still trying to figure out why she did it like that

Answer 28

Alright so solving the equation hartnn stated, what do you get for t?

\(\color{blue}{\text{Originally Posted by}}\) @hartnn
\(-
16t^2+150=0\)
\(\color{blue}{\text{End of Quote}}\)

Answer 29

i wiuld get 250 +150=0 i thinkl

Answer 30

no i divide each side by 16

Answer 31

subtract -150 from each side first.

Answer 32

What do you get?

Answer 33

16t^2=-150

Answer 34

You're missing something.

Answer 35

i am?

Answer 36

\[\color{red}{-}16t^2=-150\]

Answer 37

o that a rookie mistake

Answer 38

But it would change your t value :P

Anyway, divide both sides by -16. What do you get?

Answer 39

t^2=150/16

Answer 40

And what does 150/16 give you?

Answer 41

75/8

Answer 42

which would turn into a raidal i think. if i try to simplify it more

Answer 43

5 radical 3 / 8 i think

Answer 44

Ok that's fine, you've got \(\frac{75}{8}\). Now take the square root of that to find t. Remember, you will have both positive and negative values of time, but time is never negative , so you will have your value of time.

What do you get for your time value?

Answer 45

i would be 8.6 i think

Answer 46

\[t=\pm\sqrt{\frac{75}{8}} \ne 8.6\]

Answer 47

Use a calculator please.

Answer 48

3.0

Answer 49

Awesome, I got the same, t = 3.06 s.

Now as I mentioned earlier, velocity is the derivative of the position function. Therefore what is the derivative of \(s(t) = -16t^2+v_0t+s_0\)?

Answer 50

32+v+s

Answer 51

Remember, \(s_0\) is just a constant. In this function it represents your height, \(h=150~ft\). The derivative of a constant is 0.

So what would be your velocity function?

Answer 52

149

Answer 53

No...

Answer 54

3.06