Question

C?

Answer 1

Because i did 2.1/0.25=8.4

Answer 2

you'll have to apply the kinematics equations in both directions to find the time and horizontal distance

one moment

Answer 3

you can use equation 3 to find the time spent falling since we know:

the vertical distance (delta x or rather, delta y)
the initial velocity (which is 0 in the y-direction)
and acceleration (which is -g)

solve for t

Answer 4

0.25=0

Answer 5

2.1 would be the initial velocity right>

Answer 6

\(\color{#0cbb34}{\text{Originally Posted by}}\) @Vocaloid
you can use equation 3 to find the time spent falling since we know:

the vertical distance (delta x or rather, delta y)
the initial velocity *** (which is 0 in the y-direction) ***
and acceleration (which is -g)

solve for t
\(\color{#0cbb34}{\text{End of Quote}}\)

Answer 7

2.1 is the initial velocity in the x-direction, but we are thinking in terms of the y-direction only, so 0 is your initial velocity

Answer 8

GOt it

Answer 9

0.25=0+1/2(-98)t^2

Answer 10

-9.8*

Answer 11

good (I didn't quite get the sign convention right, so let's let

0.25 = (1/2)(9.8)t^2 so t ends up being real not imaginary
then solve for t

Answer 12

0.23

Answer 13

good

now, since t = 0.23 seconds is the time spent jumping, we use the x-direction velocity and time to calculate distance

it's just distance = rate * time = 2.1 * 0.23 = 0.483 m (we rounded up a bit so B is your answer)

Answer 14

Oh okay so me basically dividing was incorrect

Answer 15

yeah

the reason why that doesn't work is that the height is vertical and the initial velocity is horizontal, so you can't just divide distance/time

Answer 16

Got it

Answer 17

I'm probably going to log off soon, I'll probably be online tomorrow afternoon ish

Answer 18

That is fine =) Thank you