physics homework... Need help badly |
 Dec 5 2005, 01:52 AM
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#1
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 Pimp Status  Group: Member Posts: 640 Joined: Jan 2004 Member No: 1,200  |
hey guys,
This physics homework is killing me and I need some help trying to do it Equations of kinematics in two demensions and projetile motion 1) A volleyball is spiked so that it has an ititioal velocity of 15 meters per second (m/s) directed downwardat an angle of 55 degrees below the horizontal. What is the horizontal component of the balls velocity? 2)A rocket is fired at a speed of 75 meters per seond from ground level at an angle of 60 degrees above the horizontal. The rocket is fired toward an 11. 0 meter hight wall which is located 27 meters away. By how much does the rocket clear the top of the wall Nonequilibrium applications of newtons second law 1. A 1380 Kg car is moving due east with an initial speed of 27 m/s. After 8.00 seconds the car has slowed down to 17.0 m/s. Find the magnitudeand direction of the net forcethat produces the deceleration any help what so ever would be great |
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| *stephinika* |
Dec 5 2005, 02:25 AM
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#2
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Guest |
1) draw your triangle and use trig to solve.
 2) i can't think right now, and i should know this but my brain isn't working. i apologize. firstly though, draw out your triangle and use trig to split up the components. i can tell you that much for now. brain fart on the last one too. sorry. 
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Dec 5 2005, 02:46 AM
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#3
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Pimp Status Group: Member Posts: 640 Joined: Jan 2004 Member No: 1,200 |
well stephinika i thought about that too. I figured that since the angle opposite to the 55 is 35 and what i did was took cos(35)*15 should equal the length of the mystery angle. I got 12.3... Does that sound resaonable?
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| *wind&fire* |
Dec 5 2005, 02:58 AM
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#4
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[edit]acutally the persons diagram is wrong let me do that again
^and youre wrong sin35=vertical component/15 vertical component= 8.6m/s ill edit with answers 2)A rocket is fired at a speed of 75 meters per seond from ground level at an angle of 60 degrees above the horizontal. The rocket is fired toward an 11. 0 meter hight wall which is located 27 meters away. By how much does the rocket clear the top of the wall Ux= cos60 X 75 = 37.5m/s Uy= sin60 X 75 = 64.9m/s Ux is 37.5 therefore at the 25m mark the time is 1.5s therefore delta Y = Uyt + .5Ayt^2 delta Y = Uyt + .5(-9.8)t^2 delta Y = 64.9(1.5) - 4.9(1.5)^2 = 86.325 therefore since the wall if the wall is 11m tall the rocket clears it by 75.325m edit with more 1. A 1380 Kg car is moving due east with an initial speed of 27 m/s. After 8.00 seconds the car has slowed down to 17.0 m/s. Find the magnitudeand direction of the net forcethat produces the deceleration Fu=ma = 1380 X 27 = 37260N Fv = 1380 X 17 = 23460 N difference 37260 - 23460 = 13800N to the west or -13800 to the east im really not sure about this answer |
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Dec 5 2005, 08:17 AM
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#5
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Pimp Status Group: Member Posts: 640 Joined: Jan 2004 Member No: 1,200 |
hey guys i figured out the 1st and 3rd one and thanks for the help on the 2nd one wind
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| *mzkandi* |
Dec 6 2005, 02:18 AM
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#6
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^ OK, I'm assuming you need no further help, if you do pm me to reopen this.
Topic Closed. |
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