Kinematic Equations: Position & Velocity at t=2.40s

[Throwback24]

Homework Statement

Using the appropriate kinematic equations and initial values, determine the position and velocity at t=2.40 s.

(x, y)=0, 0
(vi, 0)=(25.0 m/s, 65.0 degree)

a.Horizontal Velocity of the initial velocity

b.Horizontal/Vertical Velocity at 2.40 s

c.Horizontal position at 2.40 s

Homework Equations

The Attempt at a Solution

a.(25 m/s)(Cos 65)

b.

c. X=(25 m/s)(2.40 seconds)

How do I get Horizontal/Vertical Velocity at 2.40 seconds?

 

[rigory]

someone correct me if i am wrong, but i believe that you would want to break the horizontal/vertical velocity vector into into its components and then use either trig or pythag to find the avg velocity.

 

[Throwback24]

But it's asking for the velocity @ 2.40 seconds. Don't you break up vectors into components?

 

[Gregg]

For something moving 25ms^-1 at an angle of 65 degrees to the horizontal you use. 25cos65 for the horizontal and 25sin65 for the vertical. use appropriate formula for the given times to find its velocity

 

[Throwback24]

I'm wrong on C, it has to be:

25 m/s (Cos 65)(2.40 m)

I still don't get B. Gregg, isn't that how I solve for Initial velocity?

 

[Gregg]

You can use
$$v = u + at$$ you know u, you know a and you are given t.

 

[Throwback24]

I know V is Velocity, A is Acceleration and T is time. What is U? I've never seen it.

 

[Gregg]

u is initial velocity, v₀

 

[Gregg]

V_f = V_i + at
or
V = V₀ + at