PHYSICS 111 Study Guide LAWS OF MOTION AND PROJECTILES
PHYSICS 111 Study Guide LAWS OF MOTION AND PROJECTILES PHY 111
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This 3 page Study Guide was uploaded by ntombizodwa makuyana on Saturday October 8, 2016. The Study Guide belongs to PHY 111 at Arizona State University taught by Prof Tsen in Fall 2016. Since its upload, it has received 32 views. For similar materials see General Physics in PHYSICS (PHY) at Arizona State University.
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Date Created: 10/08/16
Question: Determine the magnitude and direction of the displacement if a man walks 30km 45ᵒ north of east and then walks due east 20km. Sln: 20km 30km R 45ᵒ X direction 30cos 45+20= 41.21 Y direction 30sin 45=21.21 Resultant 2 2 0.5 R=(21.21 +41.21 ) =46 ᶿ= tan (21.21/41.21) 0 =27 o Qsn 1: A cruise ship leaving port travels 50km 45 north of west and then 70km at a heading 30 north of east. Find a) the components of the ship’s displacement vector and b) the displacement vector’s magnitude and direction. Sln: 50km 70km 45 o 30 0 X direction 70cos30 -50cos45 =25.266 Y direction 50sin45+70sin 30= 70.36N 2 2 0.5 b) R= ( 70.36 +25.266 ) =74.8 ᶿ=tan (25.266/74.8) = 70.2 north of east Projectile motion QSN: A ball is thrown so that its initial vertical and horizontal components of velocity are 40m/s and 20m/s respectively. Calculate the maximum height Sln: Using vertical components Time to reach maximum height V=u-gt T=u/g =40/9.81 =4.08 secs 2 Maximum height= ut-0.5gt =40(4.08)-0.5*9.8*4.08*4.08 =82m Qsn: A grasshopper jumps a horizontal distance of 1m from rest with an initial velocity at a 45ᵒ angle with respect to the horizontal. Find a) the initial speed of the grasshopper and b) the maximum height reached. Sln: Horizontal Displacement 1m= vcos45t ……………………………………………….1 Vertical displacement X=ut-0.5gt 2 0=Vsin45t-4.9t 2 T=0 or t=vsin45/4.9 ………………………………………………..2 Substituting 2 into 1 Vcos45Vsin45/4.9=1 V=3.13m/s b) maximum height x=0-(3.13sin45) /2*9.8 =0.250m
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