Question

a ball is dropped from rest from a tower and strikes

the ground 125m below. Approxmiately how many

seconds does it take for the ball to strike the ground

after being dropped. Neglect air resistance

Answer 1

gravity is being used in this one, right?

Answer 2

so change a with g

Answer 3

let's start
given:
Vi=0 m/s
acceleration due to gravity= 9.8 m/s^2
displacement: 125m
time=?

Answer 4

yes, you're right

Answer 5

ok

Answer 6

y=0 right?

Answer 7

and y initial=125?

Answer 8

yeah

Answer 9

so which formula are we going to use?

we don't have \(V_f\) so...?

Answer 10

I got -4.9t squared+t+125=0

Answer 11

so does this mean time for quadtatrics?

Answer 12

or did I mess up?

Answer 13

I got 5.15 s

Answer 14

is relatively close to 5.05 s, the answer

Answer 15

i want to do it, step by step , so bare with me XD
so i will use
\[\Delta d=V _{i}\Delta t + \frac{ 1 }{2 } a \Delta t ^2\]
Vi=0 so it will cancel out
it will become
\[\Delta d=\frac{ 1 }{2 } a \Delta t ^2\]
then evaluate, to solve for \(\Delta t\)
so it will be (\Delta t=\sqrt {\frac{2 \Delta d}{a}}\)

Answer 16

\(\LARGE \Delta t=\sqrt {\frac{2 \Delta d}{a}}\)

Answer 17

did you get what i did?

Answer 18

yeh

Answer 19

wat was your final answer?

Answer 20

I got time by using the quadratic formula

Answer 21

which was 5.05s, the answer

Answer 22

approx. 5.05 s

Answer 23

YEAH, my hero!!!!!

Answer 24

lowl, i'm practicing too XD
nice questions

Answer 25

ok, next problem?

Answer 26

ok! ^.^