- #1
victorhugo
- 127
- 5
The teacher does not want this homework back, I just want to learn because I love science
"1) A golf ball at a mini golf course is hit 10m NE where it hits a barrier and travels NW for another 18m. What is the balls displacement from the tee?"
square root(18^2)+(10^2)= 20.6 metres. Now for the angle I am confused, using SOH CAH TOA I always get 61 degrees, but apparently its N 15.9 W, could someone please show me why?
"5) A duck is realeased onto a lake and swims due west across it for 50 m, it then turns a swims due North for 2m. What is its displacement from its realse point?"
The displacement squared is (2^2 + 50^2) = 2504, square rooting it the displacement is 50m
For the angle,(using tan=opp./adj., opposite to the angle is 2 and adjacent ot it is 50), tan^1(2/50) = 2.3
The displacement of the duck is then 50m West 2.3° North"8) During the hammer throw at the Australian trials a hammer left the cafe traveling at 90km/h due south into a cross wind which blew it due west at 2km/h. What is its relative velocity ov the ground?"
I'm cnfused with this one, but I think that we can get it by using r^2=a^2+b^2 but making it so that the 2km/h of the wind is negative, so the square root of (-2^2) + (90^2) = 89.9m/s
"1) A golf ball at a mini golf course is hit 10m NE where it hits a barrier and travels NW for another 18m. What is the balls displacement from the tee?"
square root(18^2)+(10^2)= 20.6 metres. Now for the angle I am confused, using SOH CAH TOA I always get 61 degrees, but apparently its N 15.9 W, could someone please show me why?
"5) A duck is realeased onto a lake and swims due west across it for 50 m, it then turns a swims due North for 2m. What is its displacement from its realse point?"
The displacement squared is (2^2 + 50^2) = 2504, square rooting it the displacement is 50m
For the angle,(using tan=opp./adj., opposite to the angle is 2 and adjacent ot it is 50), tan^1(2/50) = 2.3
The displacement of the duck is then 50m West 2.3° North"8) During the hammer throw at the Australian trials a hammer left the cafe traveling at 90km/h due south into a cross wind which blew it due west at 2km/h. What is its relative velocity ov the ground?"
I'm cnfused with this one, but I think that we can get it by using r^2=a^2+b^2 but making it so that the 2km/h of the wind is negative, so the square root of (-2^2) + (90^2) = 89.9m/s