Question

Find the initial horizontal velocity, vx:
Information we know:
dy= 0.51m
dx= 1.99m
Initial vertical velocity, viy= 3.16m/s
Total time of flight, Tt= 0.64m/s

Answer 1

horizontal distance is given by speed*time \[\Large \Delta x = v_ x t \] vert is given by \[\Large \Delta y = v_ y t + 0.5 a t^2\]

Answer 2

is there acceleration in the x axis?

Answer 3

No, remember nothing accelerates it horizontally - think what force could? g only acts vertically.

Answer 4

oh. But I'm confused on which formula I should use...

Answer 5

Should I use the vf^2=vi^2+2ad formula?

Answer 6

I think we'll have to first find the angle... we can from the vertical equation... plug in all the numbers on the pic into the vert equation above: \[\Large 0.51 = v \sin \theta *0.64 + 0.5 (-9.8) (0.64)^2\] and the numbers into the horiz equation above \[\Large 1.99 = v \cos \theta *0.64\]notice we now have two equations, and two unknowns

Answer 7

if any of that is confusing... refer back to the pic and plug in the numbers and things i put on the pic

Answer 8

This is question 3. But question 4 asks for initial launch velocity and 5 asks for initial launch angle...

Answer 9

Is the angle necessary for finding the initial horizontal velocity?

Answer 10

For question 1, I found the initial vertical velocity without the angle and my teacher said I was right. Would the same apply for the initial horizontal velocity?

Answer 11

oh right, we just need vx for now... use this \[\Large \Delta x = v_ x t\] \[\Large 1.99 = v_ x*0.64\]^easy to solve for vx

\[\Large 1.99 = v \cos \theta *0.64 \] you could use that to find the angle.

Answer 12

initial launch velocity - just use pythagoras, and the vx and vy, to find the resultant v

Answer 13

oh okay. Does the initial launch angle apply to the x axis?

Answer 14

Yes, it'll be relative to it

Answer 15

Alright I understand that now but I don't really understand the initial launch velocity. How do I find the resultant v again?

Answer 16

If you draw a triangle, with vx and vy as the legs, you'll see the v is just the hypotenuse of the triangle... use Pythagoras.

Answer 17

Got it! Um if you can... can you explain these relationships to me:
Vx vs. time
Vy vs. time
Ax vs. time
Ay vs. time

Answer 18

Like if you were to sketch them on a graph.

Answer 19

Vx is constant, (no acceleration)

Vy varies with gravity... vy = viy + at so a linear graph with slope -9.8

delta x and delta y equations are above. x varies with t, y with t^2.

Answer 20

Okay. I'll try to remember all that for my physics quiz tomorrow. :D thank you so much!

Answer 21

Good luck! :)

Answer 22

thanks. (: