#1 ok
#2 I get that the player is at (-22.5,-26.8) from the net, so his shot takes the puck to (2.5,-26.8). Making it 26.9 from the net. You can figure the angle.
How did you get your solution?
#3 ok
1. Resolve an acceleration vector of 40m/s^2 [25 degrees E of S] into its components
For this I got 17 as my x component and 36 as my y component
2. A player is positioned 35m [40 degrees W of S] of the net. He shoots a puoock 25m [E] to a teammate. What second displacement does the puck have to travel in order to make it to the net?
My answer for this is 23m [S 45 degrees E]
3. Resolve a displacement vector of 5 m [E] into its components.
My answer for the x component is 5m and the y component is 0
2 answers
1. X = 40*cos25 = 36.21 m/s^2.
Y = 40*sin25 = 16.90 m/s^2.
2. d2-d1 = 25 m.
d2 = 25 + d1
d2=25 + 35[230o]=25+35*cos230+i35*sin230
= 25 - 22.50-26.81i = 2.50 - 26.81i =
27m[-84.7o] = 27m[275.3o] = 27m[5.3o E.
of S.]
3. Correct.
Y = 40*sin25 = 16.90 m/s^2.
2. d2-d1 = 25 m.
d2 = 25 + d1
d2=25 + 35[230o]=25+35*cos230+i35*sin230
= 25 - 22.50-26.81i = 2.50 - 26.81i =
27m[-84.7o] = 27m[275.3o] = 27m[5.3o E.
of S.]
3. Correct.