easy phyiscs question (help)

[foggy glen]

You let a ball weighing 230 g hang in a dynamometer. During 0.40 s you accelerate the ball upwards, and during the entire movement the dynamometer shows 3.6 N. How far does the ball travel during the movement?

[broken parrot]

Just write out the givens and determine the kinenatic equation that relates the knowns and what you want

(Assuming constant a)

[foggy glen]

the ball travels approximately 1.57 meters during the movement. this is what I got

but how do I know if I'm right or not

can u help me

@broken parrot

[broken parrot]

post your work

[foggy glen]

a=0.230kg3.6N​

@broken parrot

a≈15.65m/s2

[broken parrot]

what?

[foggy glen]

d=21​at^2

[broken parrot]

????

where the hell is 21 coming from

[foggy glen]

hold on

my bad bro

a=0.230kg/3.6N​

[broken parrot]

no

[foggy glen]

a≈15.65m/s2

[broken parrot]

that has units s^2/m

$F=ma$ solve for a.

[worthy surge]

Omegabet_

[foggy glen]

and then I used this formel:
d=ut+21​at^2

[broken parrot]

solve for a.

F=ma

what's a

[foggy glen]

15.65

m/s^2

wdy

[broken parrot]

write your work properly then

[broken parrot]

this is clearly wrong

[foggy glen]

help me bro

[broken parrot]

since that has units s^2/m

[foggy glen]

I'm so confused

[broken parrot]

Im trying, but you need to learn basic algebra first

a=F/m

since you're unable to read ig

but anyway

the kinematic quadratic reduces to $d=\frac{a}{2}t^2=\frac{Ft^2}{2m}$ (imposing $d_0=0$ and $+d$ upwards)

[worthy surge]

Omegabet_

[broken parrot]

and you know F,t and m

now it's just calculator work

[foggy glen]

so I get d≈1.252m

ig

@broken parrot

[broken parrot]

ok

[foggy glen]

is it like so?

[broken parrot]

sure

[foggy glen]

that was way easier than I thought man