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Horizontal velocity = initial v * cos (angle)
Vertical velocity = initial v * sin (angle)

horizontal= -47.62
vertical= -15.24

How long it take the shell to reach its highest point?

do you know anyone that can help? ^^

@Michele_Laino seems to be the only knowledgeable person tonight

the horizontal component of velocity is:
\[{v_x} = {v_0}\cos \theta = 50 \cdot \cos 60 = ...?\]

horizontal= -47.62
vertical= -15.24

How long it take the shell to reach its highest point?

first of all you should always start by drawing a FBD|dw:1444195069945:dw|

yup i checked it..

we can use the first formula you gave?

v2=v20+2ad

i switch it in radians mode and i still got the same value

hmm, ok i get it ^^

yup i just did it ^^

so we need to solve the distance or time first?

yeah the equation you typed out. use that.

you said that the final velocity is =0 ,
so (0)=(50)^2+2(9.8)d
d=127.55m

is that right?

so that distance is the anwer for letter c. ?

so we can use this v=v0+at to get the time?

|dw:1444196665892:dw|

|dw:1444196742365:dw|

where did you get the 55 value on the hypotenuse?

whoops. that's supposed to be a 50. this thread is long and i forgot the value.

time is required in the problem though

\[\sin60=\frac{opposite}{50}\]cross multiply to get\[50\sin60=opposite\]

its the same value as the horizontal and vertical component as it stated on required letter a.

25 and 43.3

whats next we need to figure out now?

d=v0t+12at2
v2=v20+2ad
v=v0+at
d=v0+v2t

so its clear that the time is 1 s?

0=(43.3)^2 +2 (9.8)d
d=95.658m

yup that's what i get

is that the maximum height?

How far from its firing point does the shell land?

How far from its firing point does the shell land?

solve these two then add the answers together. \[95.7=(0)t+\frac{1}{2}(9.8)t^2\]and\[0=43.3+(9.8)t\]

that will be your total time of flight.

it's been a while since i've done physics work but i'm fairly confident in this answer.