Question

what maximum height will be reached by a stone

thrown straight up with an initial speed of 35 m/s?

Answer 1

would initial speed equal v0x?

Answer 2

any tips or hints?

Answer 3

feels like this is a y equation so it is getting thrown up

Answer 4

i think since we're dealing with the y-components \(v_iy= 35 m/s\ and\ use\ V_f=0 \ m/s\) because we're looking for the maximum height

Viy=initial velocity
Vfy= final vertical velocity

Answer 5

so voy=35?

Answer 6

yep

Answer 7

0=35-9.8t

Answer 8

t=3.57

Answer 9

wait,time is not given, where did you get t=3.57?

Answer 10

well

Answer 11

from the equation Vy=Voy-gt

Answer 12

essentially 0=35-9.8t right?

Answer 13

you can use the formula \(V_f^2=V_i^2+2ad\), it's more convenient (^_^)

yes, that's right also

Answer 14

hmm

Answer 15

maximum height=y right?

Answer 16

yes

Answer 17

wait a minute, the answer is 62

Answer 18

62.5m

Answer 19

but wait

Answer 20

I have y=yo+voyt-1/2gt squared

Answer 21

which is y=o+35(3.57)-1/2(9.8)(3.57)squared

Answer 22

or am I messing up somewhere?

Answer 23

t=3.57, right?

Answer 24

voy=35?

Answer 25

it looks right to me

yes, what's wrong? O.o

Answer 26

umm

Answer 27

the answer is 62.5

Answer 28

but y=35(3.57)-1/2(9.8)(3.57)squared equals 62.5...

Answer 29

lol I made a mistake

Answer 30

DERP

Answer 31

yeah, i tried doing you're method and i got 62.49999 which is approx 62.5 also

Answer 32

*your

Answer 33

next problem?

Answer 34

how many more? O.O

Answer 35

till 10 right?

Answer 36

it is 9.24 P.M. at my place

Answer 37

yes

Answer 38

okay next one then XD