tapia jat4858 Homework 04 erskine 54935 This print out should have 33 questions Multiple choice questions may continue on the next column or page find all choices before answering 001 10 0 points Initially at time t 0 a particle is moving vertically at 5 3 m s and horizontally at 0 m s Its horizontal acceleration is 1 8 m s2 At what time will the particle be traveling at 57 with respect to the horizontal The acceleration due to gravity is 9 8 m s2 Correct answer 0 42158 s Explanation t vy0 a tan g 1 8 m s2 5 3 m s tan 57 9 8 m s2 0 42158 s 002 part 1 of 2 10 0 points During World War I the Germans had a gun called Big Bertha that was used to shell Paris The shell had an initial speed of 1 62 km s at an 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 Let vy0 5 3 m s g 9 8 m s2 vx0 0 and a 1 8 m s2 57 vyt Explanation The range R is given by and v02 R sin 2 0 g 1620 m s 2 sin 139 2 9 8 m s 175 69 km vt 57 003 part 2 of 2 10 0 points How long was it in the air Correct answer 309 675 s vxt The vertical velocity is vyt vy0 g t and the horizontal velocity is vxt vy0 a t a t The vertical component is the opposite side and the horizontal component the adjacent side to the angle so vy g t vyt 0 vxt at a t tan vy0 g t a t tan g t vy0 1 Explanation The time in the air is t R v0x R v0 cos 0 1 7569 105 m 1620 m s cos 69 5 309 675 s tan 004 10 0 points tapia jat4858 Homework 04 erskine 54935 Given The battleship and enemy ships A and B lie along a straight line Neglect air friction A battleship simultaneously fires two shells at these two enemy ships 2 Explanation Both have the same velocity at the time of release Gravitational acceleration does not change horizontal velocity so the plane will be directly above the package 006 part 2 of 4 10 0 points What is the horizontal distance from the release point to the impact point Correct answer 4795 23 m battleship A B If the shells follow the parabolic trajectories shown in the figure which ship gets hit first 1 both at the same time 2 A Explanation 1 For the vertical fall h g t2 or t 2 s 2h The horizontal distance traveled is g x vs t v 2h g 3 need more information 4 B correct 167 m s s 2 4040 m 9 8 m s2 4795 23 m Explanation The time interval for the entire projectile motion is given by ttrip trise tf all 2 trise where trise is the rising time from 0 to the maximum height and tf all the falling time from h to 0 In the absence of air resistance s 1 2 2h trise tf all h g tf all or ttrip 2 2 g So the smaller is h the smaller is ttrip In other words enemy ship B will get hit first 007 part 3 of 4 10 0 points A second package is thrown downward from the plane with a vertical speed v1 77 m s What is the magnitude of the total velocity of the package at the moment it is thrown as seen by an observer on the ground 005 part 1 of 4 10 0 points A plane is flying horizontally with speed 167 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 package hits the ground the plane will be Explanation The velocity is the vector sum of the vertical and horizontal components of velocity as seen from the ground Hence the scalar speed is q s v 2 v12 q 167 m s 2 77 m s2 1 directly above the package correct 2 ahead of the package 3 behind the package Correct answer 183 897 m s 183 897 m s 008 part 4 of 4 10 0 points What horizontal distance is traveled by this package tapia jat4858 Homework 04 erskine 54935 Correct answer 3659 37 m 2 v Explanation The time of the vertical fall is now determined by h v1 t 0 t 1 2 gt 2 4 v 1 2 g t v1 t h 2 q v1 5 v v12 4 12 g h 2 12 g q v1 v12 2 g h g 21 9124 s 6 v 7 v 8 v 9 v The horizontal distance is x vt 167 m s 21 9124 s 3659 37 m 10 m 009 10 0 points A target lies flat on the ground 5 m from the side of a building that is 10 m tall as shown below The acceleration of gravity is 10 m s2 Air resistance is negligible A student rolls a 5 kg ball off the horizontal roof of the building in the direction of the target v 10 m 3 v 5m The horizontal speed v with which the ball must leave the roof if it is to strike the target is most nearly 1 v 5 5 m s 10 v 3 5 2 m s correct 2 2 m s 5 5 m s 5 3 m s 5 5 3 m s 3 2 5 m s 5 2 m s 5 3 m s 3 5 m s Explanation m 5 kg not required h 10 m x 5 m and g 10 m s2 Observe the motion in the vertical direction only and it is a purely 1 dimension movement with a constant acceleration So the time need for the ball to hit the ground is s 2h t g and the horizontal speed should be x v t for the ball to hit the target Therefore r g v x 2h s 10 m s2 5 m 2 10 m 5 m s 2 5 2 m s 2 tapia jat4858 Homework 04 erskine 54935 010 10 0 points A 0 46 kg rock is projected from the edge of the top of a building with an initial velocity of 9 84 m s at an angle 56 above the horizontal The building is 14 5 m in height g x2 2 v0 cos 2 g x2 tan x h 0 2 v0 cos 2 x tan Since s g 2 v0 cos 2 9 8 m s2 2 9 84 m s 2 cos2 56 0 161839 m 1 a 4m 56 b tan tan 56 1 48256 14 5 m 9 8 Building 4 c h 14 …