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September 12, 2007
PHY2053
Quiz 2 (Chapter 3)
Name:
UFID:
1. (5 pts) Jacksonville is 120miles away from Gainesville and in a direction 45° north of
east. Tallahassee is 250miles away from Gainesville and in a direction 30° north of west.
Find the direction of Tallahassee relative to Jacksonville.
Jacksonville relative to Gainesville = (120cos45°, 120sin45°) = (84.8, 84.8)
Tallahassee relative to Gainesville = (250cos150°, 250sin150°) = (-217, 125)
Tallahassee relative to Jacksonville = (-217, 125) - (84.8, 84.8) = (-302, 40.2)
θ = tanˉ¹ (40.2/-302) + 180º=-7.58 + 180 = 172º, 8º north of west
2. (5 pts) A stone is thrown from the top of a building at an angle of 45° above the
horizontal. The height of the building is 80m. The stone hits the ground 100m away
from the bottom of the building. What is the angle of the velocity vector when the stone
hits the ground? (You can use g=10m/s² for this problem.)
The time when the stone reaches the ground and the initial velocity are determined by
the following simultaneous equation:
100 = v₀cos45ºt, -80 = v₀sin45ºt – (1/2)10t²
Solving the 1st equation for v₀ and substitute the result in the 2nd equation, we get
-80 = 100 - 5t², t² = 36
t = 6s, v₀ = 100/(cos45ºt) = 23.6m/s
The final velocity = (23.6cos45º, 23.6sin45º - 10×6) = (16.7, -43.3)
θ = tanˉ¹(-43.3/16.7) = -68.9º, 68.9º below the horizontal
3. (5 pts) Suppose you throw a ball against a wall with a velocity of 40m/s at an angle of
30° to the horizontal. The wall is 50m away from you. Find the height of the ball when it
hits the wall.
First find the time when the ball hits the wall:
50 = 40cos30ºt, t = 1.44s
The height at the time = 40sin30°(1.44) - (1/2)(9.8)(1.44)² = 18.6m
4. (5 pts) A river flows due east at 2.00m/s and it is 80m wide. A boat crosses the river
from the south shore to the north shore with a velocity of 10.0m/s due 30° west of north
relative to water. How far upstream or downstream is the boat when it reaches the
north shore?
We take east as +x and north as +y.
The velocity of the river relative to Earth = (2, 0)
The velocity of the boat relative to the river = (10cos120º, 10sin120º) = (-5, 8.66)
The velocity of the boat relative to Earth = (2, 0) + (-5, 8.66) = (-3, 8.66)
θ= tanˉ¹(8.66/-3) + 180 = 109.1º, 19.1º west of north
D = 80tan19.1º = 27.7m, 27.7m upstream