Question

2 questions @vocaloid

Answer 1

i know it's 4*gravity+20 but I'm doing something wrong

Answer 2

@TheSmartOne

Answer 3

any idea ?

Answer 4

a

Answer 5

given this quadratic for distance

Answer 6

but a is wrong because you don't find derivative of this one apparently or maybe you do I did

Answer 7

the variable is t and given t=4 sec

what you get than you substitute this value of t=4 inside this quadratic in place of t ?

Answer 8

@Shadow your opinion please ?

Answer 9

so is it d then

Answer 10

no 200 feet is the distance what make in 4 second

Answer 11

so d lol

Answer 12

and now you know the distance and the given velocity 20 ft/s

Answer 13

this particle when will moving to the left ?

Answer 14

yes

Answer 15

e?

Answer 16

why e.? for this value of t you get the s negativ ?

Answer 17

hope helped

Answer 18

From the equation for \(s(t)\), you should have got:

\( v(0) = \frac{ds}{dt} (0) = \color{red}{-} 14\)

\(a(0) = \frac{d^2s}{dt^2}(0) = \color{red}{+}32\)

So the start of the question is wrong --> should specify an initial **upward** velocity of 14 fps

Or the equation for \(s(t)\) is wrong .... etc etc.

Acceleration is +32fps constant downward which is the big clue

Answer 19

which question?

Answer 20

dis

Answer 21

so s it a

Answer 22

@Vocaloid

Answer 23

the question is internally inconsistent and cannot have an answer

Answer 24

is the second question e?

Answer 25

\(\frac{ds}{dt} = 180- 32 t\)

\(\frac{ds}{dt}(0) = 180 \) <-- means +ve velocity is to the right

to left means that:

\(\frac{ds}{dt} \lt 0\)

\(\implies 180- 32 t \lt 0\)

Yeet?!?!

If ya got problems from there, ya got problems :(