1. A hiker used a compass to orient himself as he walked 5.00 km in a direction 30.0 north of west, and then an additional 4.00 km in a direction 40.0 south of west. The hiker walked for 1.75 h. Calculate the magnitude and direction of the hiker's velocity of the hiker using vector components .

2. Lily is able to paddle her canoe at 2.5 m/s in still water. She drops the canoe into a river which is flowing north at 1.0 m/s. Draw sketches showing how to determine the direction of the magnitude and direction of the velocity of the canoe relative to the shore using vector addition

a. If she paddles north

b. If the paddles south

c. If he paddles east

3. Athena is planning a flight from Toronto to Montreal, which lies due east of Toronto. Her aircraft flies at a speed of 800.0 km/h in still air. A wind of 50.0 km/h is blowing from [W 30.0 N]. In what direction must she aim the airplane to fly directly to Montreal ?

4. Complete each of the questions below. Show all your steps and work.

a. An oil tanker sailing west has an engine problem. It coasts to a stop at 0.50 h, moving 10.0 km in a straight line. Calculate the average acceleration of the ship, in meters per second squared (m/s²).

b. A ball launcher on the ground is aimed straight up, and it launches a baseball with an initial velocity of 49.0 m/s [up]. The ball continues upward to a stop, and then falls until caught by a player leaning out of a window 78.4 m above the ground. How long is the ball in the air before the player catches it?

5. An archer releases an arrow at a speed of 75.0 m/s at an angle of 35.0 above the horizontal.

If the arrow struck the target at the same height above the ground as the launch point. Calculate how far the target is the archer.

please explain how you got the answer.

X wind The plane has to compensate for the wind, which is me angle theta. Break the wind into it's X and Y from which would be plane velocity 800 sin theta and 800 cos theta. We're solving for theta components. now, still breaking it into its X and Y component. X is west. X on unit circle is Cos. Y is north and is Sin on unit circle. Since it has to go directly east, that means it's north and sound and ante these velocity compensation must add up to Q. 50 wind velocity on 800cos0=800-43.3 Xax 15 cos0= 756.7/800 cos0~0.9459 50 Co5 30 - 501 - - 43.3 kirk YOU X: 5 Since sin^20+cos^20=1... sin^20=1-cos^20 50 Sin 30 =- 50 - 2 = - 25_0 km/h Sin^20~1-(0.9459)^2 =0.1057 Sinesqrt(0.1057)~0.3253 Now do Y axis 800sin0-25=0 800x0.3253-25=0 0~arcsin((25)/(800x0.3253)) ~ 1.790E = Plane velocity . wt Vo Resultant velocity = R 70 M 30' Vw = Wind velocity - - YOUVADate _ From diagram , VetV w = R = > VP = R- VN @ Vis , IVNT = 50 km/h. X- axis = 1 V w/ Cos 30' T Y- axis = 1 V w / sin 30' ( - ] ) Up /VP | : 800 km/W. X - axis = COSD Z Y-axis = | san & J. Since the resultant R has no Y-comp, that means Net Y comp. of Vw & Vp iszero. -) 50 sin 30 0 ( ] ) + 800 smy ( ]) = 0. ALT AN/ 2 ) 50 sin 30 1 12 80osmo Uv/ D) smo = 50 sing 30 sin 30 ALLINe 800 16 1 $ = sin ! son 30 2 1.7907 ans