# Projectile Motion (Displacement and Velocity vector)

hey guys, i have this quesiton im diong, i got allt he answers but theres one prob

Question:

A projectile is fired at 50 m/s and 60 degrees along a level surface.

a) Find the velocity and displacement VECTORS at t = 1.0 s

i used v2 = v1 + at
v2 = 43.3 + (-9.8) (1)
v2 = 33.5 m/s
thats the velocity at 1 second

now displacement i used
d = v1t + 1/2 at squared
i got y = 43.3 (1) + 1/2 -10 (1) squared
to get
displacement = 38.4 meters

ofcourse the y velocity i got it from : 50 sin 60 = 43.3 m/s and x velocity 50cos60 = 25 m/s as you know

but teh teacer said those aer right numbers but he ased for VECTORS so he wants 33.5 and an angle beside it i assume, i am lost now, i need help finding the displacement and velocity VECTORS :D
thanks guys!

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i posted this in the other forum too , didnt know which ones more right lol thanks, but need it soon

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Doc Al
Mentor
SS2006 said:
a) Find the velocity and displacement VECTORS at t = 1.0 s

i used v2 = v1 + at
v2 = 43.3 + (-9.8) (1)
v2 = 33.5 m/s
thats the velocity at 1 second
That's just the y-component of the velocity. You need to combine this with the x-component to get the correct velocity vector.

ok thats what i dont know how to do :( can you give me instructions, thanks brother :)

actually i think i get it

just checking

33.5 and 64.6 degrees for velocity

displacement is 38.4 and 58.4 degree

Doc Al
Mentor
SS2006 said:
just checking

33.5 and 64.6 degrees for velocity
No. As I said before, the 33.5 is just the y-component of the velocity. What's the x-component? What's the magnitude? What's the angle of the velocity vector?

i just said 64.6 degrees :D
so
33.5 m/s then theres that angle thingy and i put 64.6 degrees in it
right

Doc Al
Mentor
Answer the questions in my last post.

Doc Al said:
No. As I said before, the 33.5 is just the y-component of the velocity. What's the x-component? What's the magnitude? What's the angle of the velocity vector?
y comp: 33.5
x comp: 15.2
magnitude: you mean direction?
angle: 64.6 degrees

he said he wants the answer as a vector, can you please make this easier on me and tell me what the forma tof a vector is :D

os (x,y)?

or
33.5 then <64.6 degrees

or (x,y) then < 64.6 degrees

got to sleep soon and the asnwer is due 2morrow, thanks!

Doc Al
Mentor
SS2006 said:
y comp: 33.5
OK
x comp: 15.2
No. $v_x = 50 \cos (60)$
magnitude: you mean direction?
No, I mean magnitude. If you have the components, use the Pythagorean theorem to find the magnitude.
angle: 64.6 degrees
This is wrong since your x-comp is wrong.

he said he wants the answer as a vector, can you please make this easier on me and tell me what the forma tof a vector is :D
One way is to give a magnitude and a direction. (You can also express it as components, but I suggest magnitude and direction.)

Last edited:
isn't the X usually cos 60, not sin60?
also he wants it for the FIRST second, t = 1
so 50cos60 is initial, thats why i got 15, for t = 1 seconds

Last edited:
Doc Al
Mentor
SS2006 said:
isn't the X usually cos 60, not sin60?
also he wants it for the FIRST second, t = 1
so 50cos60 is initial, thats why i got 15, for t = 1 seconds
You are right about $v_x = 50 \cos (60)$--that's what I meant to write. But since $\cos (60) = 0.5$, $v_x = 25$, not 15.

true
i got 15 casue i did
v2 = v1 + at
so 25 - 10 (1) cause he wants 1 second
is 15, as the x displacement
did the same thing with Y and u said its correct
im confused :(

Doc Al
Mentor
Only the vertical direction is accelerated (gravity acts down, not sideways). The horizontal speed remains constant.

goddd no wonder i got 9/10
'mechanical error'
lol i wish u told me yesteryda
thanks bro
case closed :D