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PHY 9A

# PHY 9A - ’jmiﬂr—wv W— 3.73 a The acceleration is...

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Unformatted text preview: ’jmiﬂr—wv W— ____ 3.73: a) The acceleration is given as g at an angle of 53.1° to the horizontal. This is a 3—4—5 triangle, and thus, em = (3/5)g and (1,, = (4/5)g during the “boost” phase of the ﬂight. Hence this portion of the ﬂight is a straight line at an angle of 53.1° to the horizontal. After time T, the rocket is in free ﬂight, the acceleration is an, m 0 and (1,, 2 g, and; ﬁlm: franilirdi 'eqtrctérmrb 'd‘l 'prtiﬁﬁi‘ﬁlre ‘nnirinm “apply. Tr‘JLnfnrg ﬁlliib L‘Uﬂb‘tlug 'lhf'db‘b' "(f1 ‘lhm ﬂight, the trajectory is the familiar parabola. y b) During the boost phase, the velocities are: U; = (3/5)gt and 7),, = (4/5)gt, both straight lines. After t = T , the velocities are van = (3/5)gT, a horizontal line, and 2),, = (4/5)gT — g(t — T), a negatively sleping line which crosses the axis at the time of the maximum height. vx ‘5' c) To ﬁnd the maximum height of the rocket, set 1),, = 0, and solve for t, Where t = 0 when the engines are cut off, use this time in the familiar equation for y. Th2us, using t m (4/5)T and ymax = 210 + Hoyt — égt2, gm = §9T2 + égT (3T) — ég (gT) , gm 2 £93!"2 + £9312 — %9T2. Combining terms, ymax = £3ng (:1) To ﬁnd the total horizontal distance, break the problem into three parts: The boost phase, the rise to maximum, and the fall back to earth. The fall time back to earth can be found from the answer to part (c), (IS/25M?!”2 = (1/2)gt2, or t = (6/5)T. Then, multiplying these times and the velocity, 3: = \$59112 + (g-gT) (g—T) + (ggT) (gT), or :1: = %gT2 + %9T2 + 31%ng Combining terms gives :1: = 53-91”. 3.16: 3.) Solving Eq. (3.18) with y: 0, 310— — 0. 75 111 gives t— m 0.391 s. b) Assuming a horizontal tabletop, voy— = 0 and from Eq. (3. 16), 110,;— — (a: —— 2:0)/t 3.58 m/s. 0) On striking the floor, 1J3,f = —gt = — gyo = —3.83 m/s, and so the ball has velocity of magnitude 5.24 m/s, directed 469° below the horizontal. d) m) 0 0.1 0.2 0.3 0.4 1 Although not asked for in the problem, this y vs. :1: graph shows the trajectory of th: tennis ball as viewed from the side. 600111 3.26: a) horizontal motion: :3 — 3:0 2 120x33 so it = m vertical motion (take +3; to be upward): y — yo 2 ‘Ugyt + éayt2 gives 25.0 m = (no sin 43.0°)t + %(—9.80 m/sZﬁ2 Solving these two simultaneous equations for v0 and it gives v0 = 3.25 m/s and t = 2.51 s. b) vy when shell reaches cliif: ’Uy = veg + ayt = (32.6 m/s) sin43.0° — (9.80 m/sZ)(2.51 s) = —2.4 m/s The shell is traveling downward when it reaches the cliff, so it lands right at the edge of the cliff. 3.27: Take +3} to be upward. Use the vertical motion to ﬁnd the time it takes the suitcase to reach the ground: 1109 m no sin 23°, my = —9.80 m/sz, y ~— yo = —114 m, t = ? y —— yo = 1709.5 + éayt2 gives t = 9.60 s The distance the suitcase travels horizontally is :1: — x0 = '00:: = (no cos 23.0“)t = 795 m ————______________________________________________ 3.35: b) No. Only in a circle would and point to the center (See planetary motion in Chapter 12). c) Where the car is farthest from the center of the ellipse. 3.36: Repeated use of Eq. (3.33) gives a) 5.0 = m/s to the right, b) 16.0 m/s to the left, and c) 13.0 = m/s to the left. 3.37: a) The speed relative to the ground is 1.5 m/s + 1.0 m/s = 2.5 m/s, and the time is 35.0 111/25 m/s :2 14.0 s b) The speed relative to the ground is 0.5 m/s, and the time is 70 s. 3.38: The walker moves a total distance of 3.0 km at a speed of 4.0 km/h, and takes a time of three fourths of an hour (45.0 min). The boat’s speed relative to the shore is 6.8 km/ h downstream and 1.2 km/h upstream, so the total time the rower takes is 1.5 km 1.5 km m‘l— m = 1.4711]? =88 min. 3.45: a) The ex = 0 and ay = "2,6, so the velocity and the acceleration will be perpen— dicular only when 1),, = 0, which occurs at t z 0. b) The speed is v = (a2 + 462t2)1/2, ale/cit :2 0 at t = 0. (See part d below.) 0) 7' and v are perpendicular when their dot product is 0: (at)(a) + (15.0 m — ﬁt2)>< («—2313 = 02t-(30.0 m)ﬁt+2;62t3 : 0. Solve this for t: t = :k‘ / WW : +5208 5, and Os, at which times the student is at (6.25 m, 1.44 m) and (O In, 15.0 m), respectively. (1) At t 2 5.208 s, the student is 6.41 m from the origin, at an angle of 13° from the m—axis. A plot of (£06) : (\$(t)2+y(t)2)1/2 shows the minimum distance of 6.41 m at 5.208 s: 012345678910 2 e) In the 13-3; plane the student’s path is: 0-2 4-02.} B 10 12 x(r} vaﬁﬁnw'rhMmu-mmr-W wmmmumm maximum: ..... 3.54: In terms of the range R and the time t that the balloon is in the air, the car’s original distance is d = R + 'th. The time t can be expressed in terms of the range and the horizontal component of velocity, t = ”D ciao, so (1 = R (l + m). Using R = '03 sin 2050/9 and the given values yields of = 29.5 m. 3.7: a) x(t) b) “,7 = 055—266: (2.4 m/s)§ —— [(2.4 m/s2)t]5 T? I ~2ﬁf: (—2.4 111/32); c) At t = 2.0 s, the velocity is "32(24 m/s)i’ — (4.8 m/SH; the magnitude i: ”(2.4 m/s)2 + (—4.8 m/s 2 = 5.4 m/s, and the direction is arctan (if—f) = —63°. Tht acceleration is constant, with magnitude 2.4 In/s2 in the -—y—direction. d) rThe velocity vector has a component paraliel to the acceleration, so the bird is speeding up. The bird is turning toward the —y—direction, which would be to the bird’s right (taking the +2— direction to be vertical). ...
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PHY 9A - ’jmiﬂr—wv W— 3.73 a The acceleration is...

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