# How far down will the ball travel and what is the magnitude of the ball's initial vector?

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

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

Hints

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.

• for 3) I got 150ft and 4) was 85.44m/s Is this right? please let me know, thank you – 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.