Confused a little with $V_x$ and $V_y$ components and how to find the displacement of X.

A football is kicked with an initial velocity of $V_x = 30 \text{ ft/sec}$, and $V_y=80 \text{ ft/sec}$

1) How high will it reach?

My work,

$V_y =\frac {d(80t - 16t^2)}{dt} = 80 - 32t$

$80-32t = 0$

$-32t = -80$

$t= 2.5s$

$S_y(2.5) = 80(2.5) - 16(2.5^2)$

$= 100ft$

2) How long will it take?

$80-16t^2 = 0$

$8t(10 - 2t) = 0$

$-2t = -10$

$t= 5s$

3) How far down will it travel?

(This is the part I don't understand, how do you incorporate both $x$ and $y$ components. I found a velocity of $85.44$, but I'm not sure where to go off from then.)

4) What's the magnitude of the football's initial vector?

(Also do not understand how to find this part).

Thank You

  • $\begingroup$ I multiplied 30ft/sec to 5s.. got 150ft, would this be the x velocity if the x was constant? $\endgroup$ – Oninez Sep 30 '14 at 5:25
  • $\begingroup$ No it's the distance travelled sorry for the wrong notation $\endgroup$ – Jasser Sep 30 '14 at 5:37
  • 1
    $\begingroup$ Yes the answers are 150 for thirt part and 85.44 for the fourth part. $\endgroup$ – Jasser Sep 30 '14 at 5:48
  • $\begingroup$ Alright, I understand it. Thank You very much! $\endgroup$ – Oninez Sep 30 '14 at 6:33
  • $\begingroup$ You're welcome. I am glad to help. $\endgroup$ – Jasser Sep 30 '14 at 8:32


3) Treat the ball as a projectile, you can use the knowledge of the motion for a projectile to find the distance.

4) The Inital velocity is the vector sum of the initial $V_x$ and $V_y$ components of the velocity.

  • $\begingroup$ for 3) I got 150ft and 4) was 85.44m/s Is this right? please let me know, thank you $\endgroup$ – Oninez Sep 30 '14 at 5:36

$S_x=30t$ substitute t=5 for the third part.

For the initial velocity vector $V_x=30$ and V_y=80-t(0)=80$

So from here find the magnitude of the initial velocity.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.