i need help with a Physics problem?
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Maximum range is reached if the missile is launched at an angle of 45 degrees from the horizontal. This can be proved using calculus. Thus the problem can be solved using this known fact and the given range of 8500km.\n\nThe vertical component of the velocity =vsin45. Since the angle is equal to 45 degrees, the horizontal component is the same, i.e. vsin45.\n\nThe time it takes the missile to travel 8500km  can be computed using the basic formula: time=distance/velocity. Thus \n\nt =8500/vsin45  Eq. 1). Where vsin45 is the horizontal component of the initial velocity of the missile.\n\nNow use the formula: v^2-u^2=2as, and substitute known values, as follows:\n \n(vsin45)^2-0=2*9.8s Equation 2 \n\nWe take vsin45 as the final vertical component of the velocity; u=0, being the initial velocity at the top of the flight of the missile; a the acceleration of gravity equal to 9.8m/s^2, and s is the maximum altitude reached in the trajectory.\n\n(You also have to know that the initial vertical component of the velocity upward is equal to the final vertical velocity downward  of the free falling body dropped from the top of the flight of the missile.)\n\nWe now use another formula for s:\n  \ns=ut+1/2at^2\n  s=0+1/2*9.8t^2   Equation 3)\n\nBut from equaton 1) t=8500/vsin45 This is for the entire flight of the missile. The time for a free falling body to reach the ground from the top of the light of the missile is only half this. Thus divide the given expression by 2 and substitute this in  equation 3):\n\ns=1/2*9.8*[1/2*8500/v sin45]^2\n\nsubstitute the value of (vsin45)^2 given in equation 2):\n\ns=1/2*9.8*1/4*8500^2/2*9.8s\ns=1/16*72250000/s\ns^2=4,515,625   \ns=2,125 km
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here's a site with pics that might give you some ideas:\n\n\nhttp://www.physics.usyd.edu.au/competition/comp1998/comp98.html
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u can consult ur physics book
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I suppose the problem was already due so I'll finish it for you now.\n\nF=m*a\na=F/m= 2*10^6N/1.2*10^6kg = 1.67m/s^2 * g/9.81m/s^2 = 0.17g
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these are data from end fixing of the structure you should introduce to solve the system of equations, otherwise you would get infinite solutions or (rigide body motion)-Good Luck
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Try Howstuffworks.com
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Hi Mark, I will answer your question if you answer mine first.\nWhy are you asking this question, it seems like a lot of people in school are trying to take the easy way out of a physics or math problem.\nWill answer if you are not doing said above!\nCheers, Susie
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depends on the direction of the force
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try the following. they are really helpful.\n\nwww.tutor4physics.com\nwww.physicsclassroom.com
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You have to calculate p=massXspeed before the "collision".  The quantity remains constant, so if the mass increases, the speed must decreases by the same ratio for the p to stay the same.  (First, you have to convert the 8N weight to a mass using the 9,8N/kg gravity constant)
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in order to solve this problem, you need to work part (b) first before (a)\n\nb): you know gravity (g=32.2 ft/sec) acts opposite the rocket, so calculate, using:\n\nv_final = v_initial - 2*g*y_1\n\nyou know that the final velocity is zero (max height) and you are given the inital velocity. the negative sign is to account for the opposite direction of gravity. move the equation around to solve for y_1 (height from tower to max. height). then add 100 feet (to account for the tower) and you have your answer.\n\na): since you know your maximum height (from part b), substitute that answer for h. [*NOTE: remember to subtract 100 feet from your answer in part b, because t=0 at the top of the tower] You now have a quadratic equation and solve for t using the quadratic formula (you should know what that is). your answer will be the answer!\n\nThis prob is easy only because it worries upto the max height. if we had to worry about the falling down part, well thats another story.\n\n** i already worked out this prob, so if you want to ask more questions/verify your answer, just post an additional response and ill modify my question to answer it. GL!
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The problem is about conservation of angular momentum.  The tricky part is (b). To solve this use this definition of angular momentum and draw a picture to find out the value of r sin θ\n\nThe angular momentum of a particle of mass m with respect to a chosen origin is given by vector product L = r x p= mvr sin θ
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gravity and air resistance.
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A quick guess without having reviewed your math, try adding g = 9.8 m/sec^2 to your answer and see if that is the answer that is needed.
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go to a math website off yahoo. it should help.
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I am terribly afraid of the equilibrium problem.. the first one on the free response!!! well, good luck!
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I'm unclear with the way you stated the problem. Am I looking for a implicit derivative. I'll check back to solve it.