Chapter 3 Problems
1. An airplane *lies 200 km due west from city A to city B and then 300 km in the direction of 30.0° north of west from city B to city C. (a) In straight-­‐line distance, how far is city C from city A? (b) Relative to city A, in what direction is city C?
C
150. km
300. km
30.0o
260. km
a) B
200. km
A
b) 2. A man lost in a maze makes three consecutive displacements so that at the end of the walks he is right back where he started. The *irst displacement is 8.00 m westward, and the second is 13.0 m northward. Find the magnitude and direction of the third displacement.
13 m
8m
3. A person walks 25.0° north of east for 3.10 km. How far would a person walk due north and due east to arrive at the same location?
dnorth = (3.10 km)(sin(25o)) = 1.31 km deast = (3.10 km)(cos(25o)) = 2.81 km
4. A Pitcher for the Cincinnati Reds, Aroldis Chapman, has had pitches clocked at 105 miles per hour (46.9 m/s). If Chapman threw a pitch horizontally with this velocity, how far would the ball fall vertically by the time it reached home plate, 60.5 ft (18.4 m) away?
x = vxt t = x/vx t = 18.4 m / 46.9 m/s = 0.392 s ∆y=voyt + (1/2)ayt2 = 0.754 m 5. A plane aims at 30.0o North of East with an airspeed of 120 m/s. Heavy wind blows west at 40 m/s. What is the plane’s speed relative to the ground? 40 m/s
120. m/s
30.0o
104. m/s
120 m/s @ 30o N of E = 60.0 m/s north and 104 m/s east. total east = 104 m/s east + 40 m/s west = 64 m/s east
total north = 60.0 m/s
60.0 m/s
6. A brick is thrown upward from the top of a building at an angle of 25° to the horizontal and with an initial speed of 15 m/s. If the brick is in *light for 3.0 s, how tall is the building?
vx = vocosθ = 13.6 m/s v0y = vosinθ = 6.34 m/s ∆y = v0yt + (1/2)ayt2 = (6.34 m/s)(3 s) + (1/2)(-­‐9.8 m/s2)(3 s)2 = -­‐25.1 m
Building is 25.1 m tall.
7. A place kicker must kick a football from a point 36.0 m (about 39 yd) from the goal, and the ball must clear the crossbar, which is 3.05 m high. When kicked, the ball leaves the ground with a speed of 20.0 m/s at an angle of 53.0° to the horizontal. (a) By how much does the ball clear or fall short of clearing the crossbar? (b) Does the ball approach the crossbar while still rising or while falling?
vx = vocosθ = 12.0 m/s x = vxt a) ∆y = v0yt + (1/2)ayt2 = (16 m/s)(3 s) + (1/2)(-­‐9.8 m/s2)(3 s)2 = 3.9 m Ball clears bar by 0.85 m
b) vy = v0y + ayt = 16 -­‐(9.8)(3) = -­‐13.4 negative velocity, so ball is on the way down. t = x/vx v0y = vosinθ = 16.0 m/s t = 36.0 m / 12.0 m/s = 3.00 s
8. A ball is thrown at 40.0 m/s at 30.0 degrees above the horizontal from the edge of a cliff. The ball lands on the ground below 6.00 seconds later. a) How high is the cliff? b) how far from the base of the cliff does the ball land?
vx = vocosθ = 34.6 m/s v0y = vosinθ = 20.0 m/s a) ∆y = v0yt + (1/2)ayt2 = (20 m/s)(6 s) + (1/2)(-­‐9.8 m/s2)(6 s)2 = -­‐56.4 m (cliff is 56.4 m high)
b) ∆x = vxt = (34.6 m/s)(6 s) = 208 m
9. A ball is thrown at 40.0 m/s at 30.0 degrees above the horizontal from the edge of a cliff. The ball lands on the ground below 175 meters from the base of the cliff. a) How much time passes before the ball strikes the ground? b) What is the height of the cliff?
vx = vocosθ = 34.6 m/s ∆x = 175 m v0y = vosinθ = 20.0 m/s a) t = ?
∆x = vxt t = ∆x/vx = 175m/34.6 m/s = 5.05 s
b) ∆y = v0yt + (1/2)ayt2 = (20 m/s)(5.05 s) + (1/2)(-­‐9.8 m/s2)(5.05 s)2 = -­‐24.0 m (cliff is 24.0 m high)
10. A ball is thrown at 40.0 m/s at 30.0 degrees above the horizontal from the edge of a cliff. If the cliff is 50.0 m high, a) how much time does it take the ball to reach the ground? b) How far from the base of the cliff does the ball land?
vx = vocosθ = 34.6 m/s ∆y = -­‐50.0 m v0y = vosinθ = 20.0 m/s ∆y = v0yt + (1/2)ayt2 -­‐50 = (20 m/s)(t) + (1/2)(-­‐9.8 m/s2)(t)2 (quadratic)
4.9 t2 -­‐ 20 t -­‐ 50 = 0 t=
∆x = vxt = (34.6 m/s)(5.83 s) = 202 m
t = 5.83s