anonymous
  • anonymous
HELP PLEASE.. on level ground a shell is fired with an initial velocity of 50.0 m/s at 60.0 degrees above the horizontal and feels no appreciable air resistance. a.) Find the horizontal and vertical component's of the shells initial velocity. b.)How long it take the shell to reach its highest point?. c.) Find its maximum height above the ground.d.) How far from its firing point does the shell land?. e.) At its highest point,find the horizontal and vertical components of its acceleration and velocity.
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
@ganeshie8
tacotime
  • tacotime
Horizontal velocity = initial v * cos (angle) Vertical velocity = initial v * sin (angle)
anonymous
  • anonymous
horizontal= -47.62 vertical= -15.24

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anonymous
  • anonymous
How long it take the shell to reach its highest point?
tacotime
  • tacotime
I know that at the highest point the horizontal component is still 15.24 because there is no horizontal acceleration. Other than that I'm stuck, sorry.
anonymous
  • anonymous
do you know anyone that can help? ^^
tacotime
  • tacotime
@Michele_Laino seems to be the only knowledgeable person tonight
Michele_Laino
  • Michele_Laino
the horizontal component of velocity is: \[{v_x} = {v_0}\cos \theta = 50 \cdot \cos 60 = ...?\]
Michele_Laino
  • Michele_Laino
whereas the vertical component of velocity is: \[{v_y} = {v_0}\sin \theta = 50 \cdot \sin 60 = ...?\]
anonymous
  • anonymous
horizontal= -47.62 vertical= -15.24
anonymous
  • anonymous
How long it take the shell to reach its highest point?
zephyr141
  • zephyr141
are you sure you put those numbers into your calculator correctly? i mean the shell is fired 60 degrees above the horizontal. i'm getting positive numbers for my y and x directions.
zephyr141
  • zephyr141
first of all you should always start by drawing a FBD|dw:1444195069945:dw|
anonymous
  • anonymous
yup i checked it..
zephyr141
  • zephyr141
so at the highest point you have no velocity in the y direction. so by using the y component you found earlier solve for when v final is 0. you have v initial. look at the kinematic equations and find the one you can work with right now.\[d=v_0t+\frac{1}{2}at^2\]\[v^2=v_0^2+2ad\]\[v=v_0+at\]\[d=\frac{v_0+v}{2}t\] i suggest finding time first then finding the distance.
anonymous
  • anonymous
we can use the first formula you gave?
zephyr141
  • zephyr141
not yet. look at it. there's 5 variables. we have three. initial velocity, final velocity, and acceleration. we need the distance (which is what we're trying to find) and time (which can be found using another equation). look at the equations and see which one we can find time.
anonymous
  • anonymous
v2=v20+2ad
zephyr141
  • zephyr141
also you're getting those numbers because your calculator is in Radians mode, you'll want to be in degree mode to get the answers of 25 for velocity in the x-direction and 43.3 for the y direction.
anonymous
  • anonymous
i switch it in radians mode and i still got the same value
anonymous
  • anonymous
hmm, ok i get it ^^
zephyr141
  • zephyr141
nah you're in radians mode right now. that's why you're getting negative values. switch to degree mode. i mean i just checked with my CASIO fx-9750GII and confirmed it. you're in radians mode. you need to be in degree mode.
anonymous
  • anonymous
yup i just did it ^^
anonymous
  • anonymous
so we need to solve the distance or time first?
zephyr141
  • zephyr141
well. i'm looking at it now and just realized we can use the second equation and bypass my original idea of solving for time with the third equation then using that time value for the first equation and solving for distance. just use the second equation and plug in your values.
zephyr141
  • zephyr141
yeah the equation you typed out. use that.
anonymous
  • anonymous
you said that the final velocity is =0 , so (0)=(50)^2+2(9.8)d d=127.55m
anonymous
  • anonymous
is that right?
anonymous
  • anonymous
so that distance is the anwer for letter c. ?
anonymous
  • anonymous
so we can use this v=v0+at to get the time?
zephyr141
  • zephyr141
nope. you're using the initial velocity that includes both the x and y velocity components. right now we're exclusively looking at the y components. the vertical directions. that's why the question asked you to separate the initial velocity into individual components. it's given to you as traveling in two directions and we want it traveling in one. up or down. which is why we needed the cosine and the sine functions.|dw:1444196423719:dw|
zephyr141
  • zephyr141
|dw:1444196665892:dw|
zephyr141
  • zephyr141
|dw:1444196742365:dw|
anonymous
  • anonymous
where did you get the 55 value on the hypotenuse?
zephyr141
  • zephyr141
whoops. that's supposed to be a 50. this thread is long and i forgot the value.
anonymous
  • anonymous
so we need x,and y value first to get the time? or y=the highest point and x =how far its firing point does the shell land?
zephyr141
  • zephyr141
forget the time. i messed that up we don't need it. you have the x and y value already. that's what you solved using 50 sin 60 and 50cos60. look at my drawing and you'll see why we used that equation to get those numbers.
anonymous
  • anonymous
time is required in the problem though
zephyr141
  • zephyr141
look at the upper left corner. SOH CAH TOA helps remember that. see the equation for sine? and for cosine?
zephyr141
  • zephyr141
\[\sin60=\frac{opposite}{50}\]cross multiply to get\[50\sin60=opposite\]
anonymous
  • anonymous
its the same value as the horizontal and vertical component as it stated on required letter a.
anonymous
  • anonymous
25 and 43.3
zephyr141
  • zephyr141
yeah. look at it. horizontal means horizon which is left to right so naturally for us that's the x directions and vertical means up and down or put simply the y directions. we're breaking the initial velocity given to us into two components that are only in one direction. the numbers you see are the initial velocity broken up. 25 m/s is in the x direction and 43.3 m/s is in the y direction.
anonymous
  • anonymous
whats next we need to figure out now?
anonymous
  • anonymous
d=v0t+12at2 v2=v20+2ad v=v0+at d=v0+v2t
anonymous
  • anonymous
so its clear that the time is 1 s?
zephyr141
  • zephyr141
we need to find the height. remember way back at the First drawing i mentioned that at the highest point the velocity in the y direction is 0? that's why that's a known but it only works if you're only and ONLY dealing with forces in the y-direction. such as the components you first found. one of those velocities (43.3 m/s) is in the y-direction. use that as your initial velocity value and 0 for the final velocity. also don't forget acceleration. \[v^2=v_0^2+2ad\]
anonymous
  • anonymous
0=(43.3)^2 +2 (9.8)d d=95.658m
zephyr141
  • zephyr141
yup that's what i get
anonymous
  • anonymous
is that the maximum height?
zephyr141
  • zephyr141
yeah. think about it. the shell has velocity in both the x and y directions. when that shell has no velocity in the y direction then that means it reached it's max height. when you launch things at a positive angle there are two areas where an object has 0 velocity in the y direction. that's at the maximum height and at it's final resting point. we found the max height.
anonymous
  • anonymous
How far from its firing point does the shell land?
anonymous
  • anonymous
How far from its firing point does the shell land?
zephyr141
  • zephyr141
ok we're still working in the y direction. we have to find time. lets use the first equation for that. now our initial velocity is going to be 0 and we know the height we're at at this moment. \[d=v_0t+\frac{1}{2}at^2\] plug in our knowns and solve for t.
zephyr141
  • zephyr141
solve these two then add the answers together. \[95.7=(0)t+\frac{1}{2}(9.8)t^2\]and\[0=43.3+(9.8)t\]
zephyr141
  • zephyr141
that will be your total time of flight.
zephyr141
  • zephyr141
you should get 8.84 seconds. we're going to use this to find the x distance. now remember that there is no acceleration in the x direction. at least the question didn't specify there being any. now use the first equation.\[d_x=v_{o,x}t+\frac{1}{2}a_xt^2\]\[d_x=(25)(8.84)+\frac{1}{2}(0)t^2=221m\]
zephyr141
  • zephyr141
it's been a while since i've done physics work but i'm fairly confident in this answer.

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