Did I Calculate the Final Velocity and Direction of the Hockey Puck Correctly?

[homevolend]

Homework Statement

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.

Homework Equations

v_x = vcosθ
v_1y=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

v_x = vcosθ
v_1y=vcosθ

and got: v_1y=5.2 m/s
and v_x=29.5 m/s


now for next part I did:

d_x=v_x(t)
20=29.5(t)
t=0.7 sec

v_2y=v_1y+(g)(t)
v_2y=5.2+(9.81)(0.7)
v_2y=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.

 

[ehild]

v_x = vcosθ
v_1y=vcosθ

You wanted to write v_1y=vsinθ...

and got: v_1y=5.21 m/s
and v_x=29.5 m/s

Do not round too early, too much.

v_2y=v_1y+(g)(t)
v_2y=5.2+(9.81)(0.7)
v_2y=12.1 m/s

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

ehild

 

[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

 

[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