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1、1RECITATIONCHAPTER 33.12 A daring 510 N swimmer dives off a cliff with a running horizontal leap, as shown in the figure below. What must her minimum speed be just as she leaves the top of the cliff so that she will miss the ledge at the bottom, which is 1.75 m wide and 9.00 m below the top of the c
2、liff?Let downward be the y direction. .0gandyxTo find the time corresponding to the displacement y = -9.00 m, we use the following equation: 201tavyyThis is a zero launch angle problem and hence the initial velocity in the y direction v0y is zero. smyt 36.1/80.922The initial velocity is calculated a
3、s follows:xvUsing the constant acceleration equation in the x direction,smsvvtxortxtavx/29.136.7510 0020Note that the weight of the diver was not used to solve the problem.23.13 A 10,000 N car comes to a bridge during a storm and finds the bridge washed out. The 650 N driver must get to the other si
4、de, so he decides to try leaping it with his car. The side the car is on is 21.3 m above the river, while the opposite side is a mere 1.80 m above the river. The river itself is a raging torrent 61.0 m wide. (a) How fast should the car be traveling just as it leaves the cliff in order to clear the r
5、iver and land safely on the opposite side? (b) What is the speed of the car just before it lands safely on the other side?Let y be the downward direction. Also, we have .0gandyx(a) Use the information given for the y direction to find the time of flight.ssmaytttvy95.1/80.93.2120The constant accelera
6、tion equation for the x direction gives 201tatvxx.sstxv/6.395.10(b) Since the car has no acceleration in the x direction, the velocity along the x direction is constant .smvx/.0The velocity in the y direction at t = 1.995 s is calculated as follows:msgtvay /6.195./80.920 The magnitude of velocity ju
7、st before it lands issmyx/3.6233.18 A balloon carrying a basket is descending at a constant velocity of 20.0 m/s. A person in the basket throws a stone with an initial velocity of 15.0 m/s horizontally perpendicular to the path of the descending balloon, and 4.00 s later this person sees the rock st
8、rike the ground. (See the figure below). (a) How high was the balloon when the rock was thrown out? (b) How far horizontally does the rock travel before it hits the ground? (c) At the instant the rock hits the ground, how far is it from the basket?Let downward be the +y direction.(a) The height when
9、 the rock was thrown out is calculated as follows:msmsmytatvyy 1580.4/8.9210.4/.22220 (b) The displacement of the rock in the x direction at t = 4.00 s is:stvx.6./.150(c) The basket descends (20.0 m/s) (4.00 s) = 80.0 m in 4.00 s, or it is (158m - 80m) = 78.0 m above the ground when the rock reaches
10、 the ground. yxd 4.980.7.6222 43.20. A man stands on the roof of a 15.0-m-tall building and throws a rock with a velocity of magnitude 30.0 m/s at an angle of 33.0 above the horizontal. You can ignore air resistance. Calculate (a) the maximum height above the roof reached by the rock, (b) the magnit
11、ude of the velocity of the rock just before it strikes the ground, and (c) the horizontal distance from the base of the building to the point where the rock strikes the ground.Let downward be the -y direction. ./3160.sin/.30sin;/2.50.3cos/.0cos 00 smmvmmv yx yand(a) At the maximum height, the veloci
12、ty in the y direction vy is zero:Using, yavy20smyy 6.13/8.9162(b) The velocity in the x direction is constant all the time,.vx/250At the ground, we have y = -15.0 m, gives yavy202./73).15)(/80.9(2)/3.16(20 smsmsvayy The magnitude of velocity of the rock just before it strikes the ground is./64/7.3/.
13、5222 sssvyx(c) Use the vertical motion to find the time of flight:tayy0 .084/80.9.167232ssmvty The horizontal distance from the base of the building to the point where the rock strikes the ground is .stvx13.4/2.5053.28. Two archers shoot arrows in the same direction from the same place with the same
14、 initial speeds but at different angles. One shoots at 45 above the horizontal, while the other shoots at 60.0. If the arrow launched at 45 lands 225 m from the archer, how far apart are the two arrows when they land? (You can assume that the arrows start at essentially ground level.)Here, we need t
15、o calculate the range for both the arrows. To find the range we proceed as follows:Find the time t for the projectile to reach the maximum height. The time of flight t is t=2t. At the maximum height vy = 0.and hence .gtORgtvy00 gvty02Range vdyxx 00sincosinco2 OR . g02sinHere, g = 9.80 m/s2 and is the angle at which the projectile is launched.The arrow which is shot at 45 has a range of 225 m. Using this information, the initial speed at which the arrow is shot can be calculated as follows: smvORsmsmgdv /47/20590in/8.2sin 0220 Both the arrows are shot with the same initial speed. Hence we can
