Another Two Dimensional motion question

  • #26
so use x = Xo + Voxt + .5at
to find t
by setting Vox 3.60ms
 
  • #27
No no no. JoshMP is incorrect. You can find t using your horizontal velocity and displacement.
 
  • #28
38
0
There is no horizontal acceleration. So Xf= Xi + Vot

Once you have t, plug it into Yf= Yi +.5(-g)t^2 and solve for Yf.
 
  • #29
ok, than in the x = Xo + Voxt + .5at^2
i know that a = -9.8
and t = .5 cuz i did V = t/D
is Xo = o and Vox = 0?
 
  • #30
38
0
"so would the final velocity be 3.60 m/s"

That was the question I was answering earlier when I said to use trig. You need to know the y-component to calculate the final velocity. I was not talking about finding t.
 
  • #31
v = d/t, you've got it the wrong way around.

And no, x = 0. But initial velocity is zero.
 
  • #32
joshMp how would you use trig if you only have one number and no degree?
 
  • #33
38
0
You have the x component, which was given to you. Find the y component using a kinematic equation. You now have the x component and the y component of the final velocity vector. Vf^2= Vx^2 + Vy^2
 
  • #34
"so would the final velocity be 3.60 m/s"

That was the question I was answering earlier when I said to use trig. You need to know the y-component to calculate the final velocity. I was not talking about finding t.
^ You need the final velocity to use that equation. Find the time to reach horizontal displacement, then use that time to find the vertical displacement.
My apologies, I thought the initial post (the one I've quoted second) was in response to v=d/t.
 
  • #35
i'm confused, so in Vf^2 = Vx^2 + Vy^2
and you said that final is 3.60 m/s than what is Vx?
 
  • #36
38
0
No. Vx= 3.6 m/s. You need to find Vy before you can find Vfinal.
 
  • #37
oh, than to find Vy = Voy - at?
 
  • #38
38
0
Yep.
 
  • #39
than what is the time and Voy
i know that a = -9.8
 
  • #40
drizzle
Gold Member
366
57
than what is the time and Voy
i know that a = -9.8

Vo is the initial velocity, you have the initial speed [17.0 m/s] and the angle [35.0° above the horizontal] and a hint:


the sine helps you solve for the vertical component.
 

Related Threads on Another Two Dimensional motion question

  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
4
Views
3K
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
6
Views
11K
Replies
4
Views
7K
  • Last Post
Replies
6
Views
2K
  • Last Post
Replies
8
Views
660
  • Last Post
Replies
5
Views
2K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
10
Views
1K
Top