# Kinematics in Two Dimensions

## 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.

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!

Related Introductory Physics Homework Help News on Phys.org
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 $$v_{final} = v_{initial} + at$$

solve for v final y component.
Then use pythagoras to find total speed (combining x and y components).

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 :(

Well now you have y velocity and the x velocity so you want to combine them to total velocity.

$$\sqrt{-4.3^2 + 14.1^2}$$

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.

Oh ok, yeah it is the hypotenuse.