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Homework
2 |
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1. Walker2 3.P.002. [235466] You are driving up a long inclined road. After 1.3 miles you notice that signs along the roadside indicate that your elevation has increased by 540 ft. (a) What is the angle of the road above the horizontal? Distance
traveled = hypoteneuse of right triangle = 1.3 miles * 5280 ft/mile = 6864 ft Sin(theta)
= elevation / hypoteneuse = (540 ft) / (6864 ft) = 0.0787 Theta =
asin(0.0787) Similar
triangles: (150 ft / length) = (540
ft) / (6864 ft) = 0.0787 Length =
(150 ft)/0.0787 = 1907 ft |
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2. Walker2 3.P.004. [235468] (a) Find the x and y components of a position vector, r, of magnitude r = 51 m, if its angle relative to the +x axis is 10.0°. x = |r| cos(10.0°) = (51 m) cos(10.0°) y = |r| sin(10.0°) = (51 m)
sin(10.0deg) |
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3. Walker2
3.P.006. [235469] A lighthouse that rises h = 42 ft above the surface of the water sits on a
rocky cliff that extends 19 ft from its base, as shown in Figure 3-32. A
sailor on the deck of a ship sights the top of the lighthouse at an angle of
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4. Walker2 3.P.022. [235479] A vector has a magnitude of 2.2 m and points in a direction that is 160° counterclockwise from the x axis. Find the x and y components of this vector. Ax = (2.2m) cos(160 deg) Ay = (2.2m) sin(160 deg)
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5. Walker2 4.P.003. [235950] Starting from rest, a car accelerates at 2.2 m/s2 up a hill that is inclined 5.5° above the horizontal. How far horizontally and vertically has the car traveled in 12 s? distance = (1/2) a t2 = (0.5)(2.2 m/s2) (12s)2 = 158.4 m Horizontal = (158.4 m) * cos(5.5deg) Vertical = (158.4
m) *sin(5.5deg) |
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6. Walker2 4.P.012. [235956] Playing shortstop, you pick up a ground ball and throw it to second base. The ball is thrown horizontally, with a speed of 26 m/s, directly toward point A (Figure 4-15). When the ball reaches the second baseman 0.55 s later, it is caught at point B. (Neglect air resistance.)
(a) How far were you from the second baseman? X = vx
t = (26 m/s) (0.55s) = 14.3 m Y –y0 =
-(1/2)gt2 |y-y0|
= (0.5)*(9.8 m/s2) (0.55s)2 = 1.48 m |
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7. Walker2 4.P.026. [235964] A soccer ball is kicked with a speed of 11.50 m/s at an angle of 26.0° above the horizontal. If the ball lands at the same level from which it was kicked, how long was it in the air? (Neglect air resistance.) 0 = y = y0 + v0y t - (1/2) g t2
= 0+v0 sin(q)t - (1/2) g t2 0 = t [v0 sin(q) - (1/2) g t] t=0 or 0 = v0 sin(q) - (1/2) g t t = 2 v0 sin(q) / g =
2.*(11.50 m/s) sin(26.0)/(9.8m/s2) [1.03] s |
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8. Walker2 4.P.033. [235966] The "hang time" of a punt is measured to be 5.00 s. If the ball was kicked at an angle of 73.0° above the horizontal and was caught at the same level from which it was kicked, what was its initial speed? (Neglect air resistance.) From question 7, Hang time t = 2 v0 sin(q) / g v0 = g t / [ 2
sin(q) ] = (9.8m/s2)(5.00s) / [ 2
sin(73.0)] [25.6] m/s |
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9. Walker2 4.P.037. [235969] On a hot summer day a young girl swings on a rope above the local swimming hole (Figure 4-20). When she lets go of the rope her initial velocity is 2.05 m/s at an angle of 35.0° above the horizontal. If she is in flight for 1.10 s, how high above the water was she when she let go of the rope? v0 = 2.05 m/s v0x = v0 cos(35deg) = 1.679 m/s v0y = v0 sin(35deg) = 1.176 m/s y = y0 +v0y t – (1/2) g t2 y-y0 = -4.641 m y=0 y0 = +4.641 m
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