Question

KINEMATICS HELP!!
2ND ATTACHMENT IS POSTED AT THE END!!

Answer 1

@ash2326 @ankit042 @bahrom7893 @bii17 @countonme123 @CarlosGP @CarolineStar @Dahlioz @dumbcow @Frostbite @skullpatrol @sam801 @StudyMathlol @Mimi_x3

Answer 2

@dpasingh @Floridagirl10 @uri @TURITW

Answer 3

@Preetha

Answer 4

Check this one if u have the final answer, i am not sure
v_y_at_max_height = u_y -gt
v_y_at_max_height = u*sin37 - 9.8*t
but v_at_max_height is zero when it reaches its maximum height
so
0=50m/s * 0.6 - 9.8m/s^2*t
9.8m/s^2*t= 50m/s
t= (50m/s)/9.8m/s^2
t = 5.10 sec (approximately)

Answer 5

Its 4 sec/....

Answer 6

According to key.But u came close.....Pls Explain ur method!!

Answer 7

U havent multiplie with 0.6

Answer 8

*multiplied

Answer 9

i think i made a littile mistake

Answer 10

yea i forgot to multiple 50 with 0.6

Answer 11

lol

Answer 12

there is the solution

Answer 13

how do u get s^2??wat is s^2??

Answer 14

Oh m/s^2 sry

Answer 15

v_y_at_max_height = u_y -gt
v_y_at_max_height = u*sin53 - 9.8*t
but v_at_max_height is zero when it reaches its maximum height
so
0=50m/s * 0.8 - 9.8m/s^2*t
9.8m/s^2*t= 40m/s
t= (40m/s)/9.8m/s^2
t = 4.08 sec (approximately)

Answer 16

Why 53??

Answer 17

the unite for acceleration is meter per second square(m/s^2)

Answer 18

Ya I figured that but why 53?

Answer 19

because we take the inclination from the above

Answer 20

@sam801 , angle is 53 not 37

Answer 21

which is 90 -37

Answer 22

oh lol you changed it

Answer 23

I know that but y is it 53??

Answer 24

@dumbcow : nice representaion

Answer 25

why 53?? Just to bring it to proper coordinate system???

Answer 26

yes in a way , you need the angle the particle makes with horizontal

the line given that has angle of 37 is not direction particle is leaving at

Answer 27

Thanks to @sam801 and @dumbcow !!!
Your efforts are much appreciated!!

Answer 28

the reason is , since we need to take the angle with respect to the x-axis(as reference) or can take y_axis, now the object is project at 90 degree but already the surface is inclined by 37 so we have to subtract the 37 from 90 to have the exact inclination.

Answer 29

got that

Answer 30

Hot it is clear now

Answer 31

hope sorry

Answer 32

It sure is!! I was kinda stuck due to 37 degrees!!Now I know that we have to shift the figure in such problems!!

Answer 33

Ta again!!:)))

Answer 34

u welcome

Answer 35

Hey another one??

Answer 36

help

Answer 37

@sam801 @dumbcow

Answer 38

I get 3.4...Is that r8??

Answer 39

no I dont get it.Calc Error

Answer 40

v is velocity of boat
F is net velocity and direction of boat with current

\[F \cos(60) = 2\sqrt{3}\]
\[F = 4\sqrt{3}\]
\[v = F \sin(60) = 6\]