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Examples in Chapter 3

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Problem 3.23 A man stands on the roof of a 150 m tall building and throws a rock with a velocity of 30 m/s at an angle of 33 0 above the horizontal. Ignore air resistance. Calculate: a) The maximum height above the roof reached by the rock b) The magnitude of the velocity of the rock just before it strikes the ground c) The horizontal distance from the base of the building to the point where the rock strikes the ground.

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Step 1: Draw It! 150 m 30 m/s 33 0 Range, R Height, H

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What do we know? The x- and y-components of the initial velocity V x0 =30*cos(33 0 )=25.16 m/s V y0 =30*sin(33 0 )=16.33 m/s The acceleration in the y-direction: a y =-g=-9.8 The acceleration in the x-direction: a x =0 The initial height of the rock, 150 m, = y 0 The initial horizontal position of the rock, 0

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Motion in the x- and y-directions

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At the top of the arc, v y (t)=0

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The magnitude of the velocity before it strikes ground. The rock strikes ground when y(t)=0

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The range of the rock The rock strikes ground after 4.111 s

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Problem 3.31 In a test of a g-suit a volunteer is rotated in horizontal circle of radius 7.0 m. What is the period of rotation at which the centripetal acceleration has a magnitude of a) 3.0 g? b) 10.0 g?

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Step 1: Draw It! 7.0 m a rad =3 g or 10 g

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What do we know? a rad =3 g or 10 g R= 7.0 m Need to find T

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Plugging and Chugging a rad =3 g or 10 g R= 7.0 m Need to find T

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Problem 3.37 A moving sidewalk in an airport moves at 1 m/s and is 35.0 m long. If a women steps on one end and walks at 1.5 m/s relative to the moving sidewalk, how much time does she require to reach the opposite side if a) She walks in the same direction as the sidewalk is moving? b) She walks against the motion of the sidewalk?

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Step 1: Draw It 1 m/s 1.5 m/s 1 m/s 1.5 m/s

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Must find relative velocity Call the slide walk as reference frame A, therefore the womans velocity in this frame v A is 1.5 m/s Call a stationary observer frame of reference, B and the slidewalks velocity in this frame is v B =1.0 m/s The end points are fixed in reference B so I must adjust the womans velocity to reference B

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Two different velocities If the woman and slide are in the same direction: V p/B =V p/A +V A/B V p/A = velocity of woman relative to slidewalk=1.5 m/s V A/B = velocity of slidewalk relative to frame B=1.0 m/s V p/B =1+1.5=2.5 m/s

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Two different velocities contd If the woman and slide are in the opposite directions: V p/B =V p/A +V A/B V p/A = velocity of woman relative to slidewalk=-1.5 m/s V A/B = velocity of slidewalk relative to frame B=1.0 m/s V p/B =1-1.5=-0.5 m/s

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Finally, In the same direction: v p/B =d/t where d=35 m and V p/B =2.5 m/s t=35/2.5=14 s In the opposite direction: v p/B =d/t where d=35 m and V p/B =0.5 m/s t=35/0.5=70 s

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Problem 3.58 A baseball thrown at an angle of 60 0 above the horizontal strikes a building 18 m away at a point 8 m above the point it is thrown. Ignore air resistance. a) Find the magnitude of the initial velocity of the baseball ( the velocity with which the baseball is thrown) b) Find the magnitude and direction of the velocity just before it strikes the building.

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Step 1: Draw It! 8 m v 0 m/s 60 0 18 m

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The Secret Weapon: The Trajectory Equation You can go through a lot of rigmarole but the most powerful tool in your projectile arsenal is this little formula below

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You know x=18 m y= 8 m = 60 0 You need to find v 0

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Part B) x=18 m, x 0 =0 y= 8 m, y 0 =0 = 60 0 v 0 =16.55 m/s and v 0x =16.55*cos(60 0 ) =8.275, v oy =16.55*sin(60 0 ) =14.33 a x =0, a y =-9.8 m/s 2 Need to find v x (t) and v y (t) when x=18 m

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