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Kinematics in Two Dimensions

  • #1

Homework Statement


A soccer player kicks the ball toward a goal that is 16.8 m in front of him. The ball leaves his foot at a speed of 16.0 m/s and an angle of 28.0 degrees above the ground. Find the speed of the ball when the goalie catches it in front of the net.


Homework Equations


x = vt
x = Vot+(1/2)at2
Vf2=Vo2+ax

The Attempt at a Solution


Vx = (cos 28)(16.0) = 14.1 m/s
Vy = (sin 28) (16.0) = 7.5 m/s

Horizontal
x = 16.8 m
V = 14.1 m/s

Vertical
a = -9.8 m/s2
Vyo = 0 m/s
(V = 7.5 m/s?)

I'm stuck as to where I should go from there or even if I started out correctly.
If you can assist me, I would be very grateful. Thank you!
 

Answers and Replies

  • #2
325
0
Yes, you have started well. First you want to find how long the ball is in the air for.
There is no acceleration in the x plane so the x speed will remain the same.
So you can find the time using only the x parameters.

However the final speed of the ball will depend on x and y velocities.
You will now have:

t
ay
vyinitial

use [tex] v_{final} = v_{initial} + at [/tex]

solve for v final y component.
Then use pythagoras to find total speed (combining x and y components).
 
  • #3
Thank you for replying.

I'm still a bit puzzled though, but if I understood correctly, this is what I did:

x = vt
(16.8 m) = (14.1 m/s)t
t=16.8 m/14.1
t= 1.2 s

Vf2 = Vo + at
Vf = (7.5 m/s) + (-9.8 m/s2)(1.2 s)
Vf = -4.3 m/s

I don't understand what to use pythagoreas for though.

Please bear with me. I'm not very good at word problems :(
 
  • #4
325
0
Well now you have y velocity and the x velocity so you want to combine them to total velocity.

[tex] \sqrt{-4.3^2 + 14.1^2} [/tex]

it makes a right angled triangle when you think about it, because x and y are perpendicular to each other, so the line that joins them (total velocity) is the hypotenuse.
 
  • #5
Oh ok, yeah it is the hypotenuse.

Thank you for your help!
 

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