Question

a ball is shot straight up from the surface of the

earth with an initial speed of 19.6 m/s. Neglect

any effects due to air resistance.

29. What is the magnitude of the ball's displacement

from the starting point after 1.00 has elapsed?

30. what maximum height will the ball reach?

31. how much time elapses between the throwing

of the ball and its return to the original launch

point?

Answer 1

voy=19.6 cause we are dealing with y, right?

Answer 2

1.00 second has elapsed for problem 29

Answer 3

for number 19 is it seconds? 1.00seconds?

and yes, we're dealing with y

Answer 4

oh okay, i just asked that XD

Answer 5

my vault lol

Answer 6

so the magnitude is v y final?

Answer 7

oh no, it is 14.7

Answer 8

where did I go wrong

Answer 9

unless if magnitude is voy-gt/2

Answer 10

though I am not sure

Answer 11

do you want to solve it here or i can just solve it on my own and just tell me what you got.?

we don't need to find for vfy for number 29, just use d=viyt+1/2at^2
viy (initial velocity is 19.6 m/s)

Answer 12

no gravity either?

Answer 13

acceleration=gravity

Answer 14

so a=9.8?

Answer 15

yes

Answer 16

if I am not mistaken

Answer 17

hmm

Answer 18

so what did you get for 29?

Answer 19

I get d=19.6t+1/2at squared

Answer 20

a=9.8 right?

Answer 21

19.6+24.01=43.61 but that isn't the correct answer

Answer 22

remeber t=1.00 sec and yes a=9.8m/s^2 and vi=19.6 m/s

Answer 23

the correct answer is 14.7

Answer 24

so I am messing up somewhere

Answer 25

ahh 4.9 and 19.6

Answer 26

I made a carless mistake

Answer 27

so 19.6-4.9=14.7

Answer 28

because velocity is up and acceleration due to gravity is down
so it will be 19.6-4.9=14.7

Answer 29

and you're right again XD
i guess i need to practice typing fast O.O

Answer 30

ahh I get it

Answer 31

I learn the most when I practice with others than practicing alone personally

Answer 32

what maximum height will the ball reach?

Answer 33

so Vy=Voy-2g(y-yo)

Answer 34

opps, thats not wright...

Answer 35

let's list what we have so far
a= -9.8 m/s^2
vi= 19.6 m/s

so for number 30, Vf will be 0 m/s
for easier calculation, use \(\Large V_f^2=V_i^2 + 2ad\)

Answer 36

30 is 19.6 hmm...

Answer 37

what a minute, the ball was shot in the air for 1 second

Answer 38

19.6 times 1=19.6

Answer 39

so the maximum height the ball will reach is 19.6 at 1 second, right?

Answer 40

or am I approaching it the wrong way?

Answer 41

we're not sure if the ball reached the maximum height in 1 second, so it's better not to use the time

Answer 42

for number 29, they just asked where will the ball after 1 sec..

Answer 43

so the ball will reach a maximum height

Answer 44

of 19.6

Answer 45

that's the initial velocity

Answer 46

the maximum height is apparently 19.6m as well according to the answer key

Answer 47

maybe it is just coincidence?

Answer 48

anyways

Answer 49

we got to wrap this up soon

Answer 50

how much time elapses between the throwing of the ball and its return to the

original launch point?

Answer 51

oh you already solved it , wait give me a sec, i'll solve it too , sorry :/

Answer 52

its fine

Answer 53

yeah, i got the same thing 19.6

okay next question XD
how much time elapses between the throwing of the ball and its return to the original launch point?
d=0
so we can use d=vit + 1/2 gt^2
now we need to use the quadratic formula ^.^

Answer 54

still there? O.o

Answer 55

yeah

Answer 56

so 4.9t squared+19.6t-14.7=0?

Answer 57

don't forget the direction
the gravity is down, velocity is up
set up as positive, gravity will be negative

Answer 58

-4.9tsquared+19.6t-14.7?

Answer 59

where did you get 14.7? it's the maximum height right?
we actually dont need it because it's asking for the time when the object came back to its original position so it will be zero: \(-4.9t^2+19.6t+0\)

Answer 60

ahh

Answer 61

derp

Answer 62

ok

Answer 63

got it

Answer 64

thank so much for all the help

Answer 65

9: 57 pm here
3 mins more , anything else?? ^.^

good luck on your exams!! you can do it!!
remember your directions (positive or negative), your units

Answer 66

when do you know when something is positive and when something is negative?

Answer 67

it depends on which direction you set as positive or negative

Answer 68

shot a rocket up

Answer 69

between shooting a rocket down

Answer 70

if you set up as positive , down will be negative and vice versa

for objects going up, it's more efficient to use \(up\) as positive
for objects thrown downwards, it's more efficient to use \(down\) as positive

Answer 71

thank man, I really do appreciate it

Answer 72

no problem (^_^)
i'm doing physics too ! so no worries ;p