tapia (jat4858) – Homework 04 – erskine – (54935) 1This print-out should have 33 questions.Multiple-choice questions may continue onthe next column or page – find all choicesbefore answering.001 10.0 pointsInitially (at time t = 0) a particl e is mov-ing vertically at 5.3 m/s and horizontally at0 m/s. Its horizontal acceleration is 1.8 m/s2.At what time will the particle be travelingat 57◦with respect to the horizontal? Theacceleration due to gravity is 9.8 m/s2.Correct answer: 0.42158 s.Explanation:Let : vy0= 5.3 m/s ,g = 9.8 m/s2,vx0= 0 , anda = 1.8 m/s2, andθ = 57◦.vxtvytvt57◦The vertical velocity isvyt= vy0− g tand the horizontal velocity isvxt= vy0+ a t = a t .The vertical component is the opposite sideand the horizontal component the adjacentside to the angle, sotan θ =vytvxt=vy0− g ta ta t tan θ = vy0− g ta t tan θ + g t = vy0t =vy0a tan θ + g=5.3 m/s(1.8 m/s2) tan(57◦) + 9.8 m/s2=0.42158 s .002 (part 1 of 2) 10.0 pointsDuring World War I, the Germans had a guncalled Big Bert ha that was used to shell Paris.The shell had an i ni tial speed of 1.62 km/s atan initial inclination of 69.5◦to the horizontal.The acceleration of gravity is 9.8 m/s2.How far away did the shell hit?Correct answer: 175.69 km.Explanation:The range R is given byR =v20gsin(2 θ0)=(1620 m/s)29.8 m/s2sin 139◦= 175.69 km .003 (part 2 of 2) 10.0 pointsHow long wa s it in the air?Correct answer: 309.675 s.Explanation:The time in the air ist =Rv0x=Rv0cos θ0=1.7569 × 105m(1620 m/s) cos 69.5◦= 309.675 s .004 10.0 pointstapia (jat4858) – Homework 04 – erskine – (54935) 2Given: The battleship and enemy ships Aand B lie along a straight line. Neglect airfriction.A battleship simultaneously fires two shellsat these two enemy ships.battleshipA BIf the shells follow the parabolic trajectoriesshown in t he figure, which ship gets hit first?1. both at the same time2. A3. need more information4. B correctExplanation:The time interval for the entir e projectilemotion is given by ttrip= trise+tfall= 2 trise,where triseis the rising time from 0 to themaximum heig ht, and tfallthe falling t imefrom h to 0. In the absence of air resistancetrise= tfall, h =12g t2fall, or ttrip= 2s2 hg.So the smaller is h, the smaller is ttrip. Inother words, enemy ship B will get hit first.005 (part 1 of 4) 10.0 pointsA plane is flying horizontally with speed167 m/ s at a height 4040 m above the ground,when a package is dropped from the plane.The acceleration of gravity is 9.8 m/s2.Neglecting air resistance, when the packagehits the ground, the plane will be1. directly above the package. correct2. ahead of the package.3. behind the package.Explanation:Both have the same velocity at the time ofrelease. Grav itational acceleration does notchange horizontal velocity, so the plane wil lbe directly above the package.006 (part 2 of 4) 10.0 pointsWhat is the horizontal distance from the re-lease point to the impact point?Correct answer: 4795.23 m.Explanation:For the vertical fall, h =12g t2, or t =s2 hg. The horizonta l distance traveled isx = v t= vs2 hg= (167 m/s)s2 (4040 m)9.8 m/s2= 4795.23 m .007 (part 3 of 4) 10.0 pointsA second package is thrown downward fromthe plane with a vertical speed v1= 77 m/s.What is the magnitude of the total velocityof the package at the moment it is thrown asseen by an observer on the ground?Correct answer: 183.897 m/s.Explanation:The velocity is the vector sum of the verticaland horizontal component s of velocity as seenfrom the ground. Hence the scalar speed i ss =qv2+ v21=q(167 m/s)2+ 77 m/s2= 183.897 m/s .008 (part 4 of 4) 10.0 pointsWhat horizontal distance is traveled by thispackage?tapia (jat4858) – Homework 04 – erskine – (54935) 3Correct answer: 3659.37 m.Explanation:The time of the vertical fall is now deter-mined byh = v1t +12g t20 =12g t2+ v1t − ht =−v1+qv21+ 4 (12) g h212g=−v1+qv21+ 2 g hg= 21.9124 s .The horizonta l distance isx = v t= (167 m/s) (21.912 4 s)= 3659.37 m .009 10.0 pointsA target lies flat on the ground 5 m from theside of a building that is 10 m tall, as shownbelow.The acceleration of gravity is 10 m/s2. Airresistance is negligible.A student rolls a 5 kg ball off the horizontalroof of the building in the direction of thetarget.5 m10 mv10 mThe horizontal speed v with which the ballmust leave the r oof if it is to strike the t argetis most nearly1. v = 5√5 m/s.2. v =5√22m/s. correct3. v =√25m/s.4. v =√55m/s.5. v =√35m/s.6. v =5√33m/s.7. v = 2√5 m/s.8. v = 5√2 m/s.9. v = 5√3 m/s.10. v = 3√5 m/s.Explanation:m = 5 kg , not requiredh = 10 m ,x = 5 m , andg = 10 m/s2.Observe the motion in the vertical directiononly and it is a purely 1-dimension movementwith a constant acceleration. So the timeneed for the ball to hit the ground ist =s2 hgand the horizontal speed should bev =xtfor the ball to hit the target. Thereforev = xrg2 h= (5 m)s10 m/s22 (10 m)=5√2m/s=5√22m/s .tapia (jat4858) – Homework 04 – erskine – (54935) 4010 10.0 pointsA 0.46 kg rock is projected from the edge ofthe top of a buil di ng with an initial velocity of9.84 m/s at an angle 56◦above the horizontal.The building is 14.5 m in height.Buildingx56◦9.84 m/s14.5 mAt what horizontal distance, x, from thebase of the building will the rock strike theground? Assume the ground is level andthat the side of the building is vertical. Theacceleration of gravity is 9.8 m/s2.Correct answer: 15.0958 m.Explanation:Let : θ = 56◦,v0= 9.84 m/s ,h = 14.5 m , andm = 0.46 kg .The flying time can be determined byx = v0xtt =xv0x=xv0cos θ.From the point where the rock was projected(set to be the origin O), the y-coordinate ofthe point where the rock struck the ground ish = v0yt −12g t2= v0sin θxv0cos θ−12gxv0cos θ2= x tan θ −g x22 (v0cos θ)2g2 (v0cos θ)2x2− (tan θ) x − h = 0Sincea =g2 (v0cos θ)2=9.8 m/s22 (9.84 m/s)2cos256◦= 0.161839 m−1,b = −tan θ = −tan 56◦= −1.48256 ,c = −h = −14.5 m , andd = b2− 4 a c= (−1.48256)2− 4 (0.161839 m−1) (−14.5 m)= 11.5846 ,x =−b ±√b2− 4 a c2 a= −−1.482562 (0.161839 m−1)±√11.58462 (0.161839 m−1)=15.0958 m .011 (part 1 of 2) 10.0 pointsConsider the setup of “aiming a gun at atarget” shown in the sketch where t he initialspeed of the bullet v0= 89.2 m/s , θ = 29.1◦and ℓ = 20.7 m .v0θℓhxyOPtapia (jat4858) – Homework 04 – erskine – (54935) 5Consider two experiment s:1) Experi ment 1 is done on the surface of theEarth, and2) experiment 2