# Homework Help: Finding components of velocity

1. Sep 28, 2011

### homevolend

1. The problem statement, all variables and given/known data

A hockey player shoots hockey puck towards net. The net is 20 metres infront of the player. If the puck leaves hockey stick at 30 m/s and 10° above ice. Find x and y components of the velocity. If the goalies catches puck just infront of net calculate final velocity of the puck with magnitude and direction included in answer.

2. Relevant equations

vx = vcosθ
v1y=vcosθ

3. The attempt at a solution[/b

Ok now for the first I assume it means find x and y components of the velocity means of the initial velocity.

so I did the above equations of

vx = vcosθ
v1y=vcosθ

and got: v1y=5.2 m/s
and vx=29.5 m/s

now for next part I did:

dx=vx(t)
20=29.5(t)
t=0.7 sec

v2y=v1y+(g)(t)
v2y=5.2+(9.81)(0.7)
v2y=12.1 m/s

v2=√ 29.52+12.12
v2=31.9 m/s

tan-1(12.1/29.5)=θ
θ=22.3°

final velocity of the puck is 31.9 m/s [22.3° Forward]

Im just wondering if I did this question correctly and if my direction is correct for the final part.

2. Sep 29, 2011

### ehild

You wanted to write v1y=vsinθ...

Do not round too early, too much.

Reconsider the sign of v1y and g. One is up, the other is down. The result you got is wrong.

ehild

3. Sep 29, 2011

### homevolend

Ok re did and got this for the x and y components

vx = vcosθ
v1y=vsinθ

v1y=5.21 m/s
and vx=29.54 m/s

now for next part I did:

dx=vx(t)
20=29.54(t)
t=0.677 sec

v2y=v1y+(g)(t)
v2y=5.21+(-9.81)(0.677)
v2y= -1.4 m/s

v2=√ 29.542+(-1.4)2
v2=29.5 m/s

tan-1(-1.4/29.54)=θ
θ= what did I do wrong?

not sure what I did wrong

4. Sep 29, 2011

### ehild

Do not round off during calculations. Keep at least 4 significant digits. The final angle will be negative as the puck is falling, its vertical component of velocity is negative.

ehild