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 October 5th, 2007, 06:31 PM #1 Senior Member   Joined: Apr 2007 Posts: 2,140 Thanks: 0 Physics Homework Objects in free fall near the surface of the Earth have a constant acceleration of 9.8 m/s². (Here are some of the questions that was included in my homework) 1. A stone dropped from the top of a building strikes the ground in 4.2 seconds. How tall is the building? 2. A bomb dropped from an airplane is given an initial downward speed of 10.4 m/s. What is its final velocity if it hits the ground 40.3 seconds after being released? Can someone tell me how to solve these questions, using the constant acceleration on top? Also, show the equations that you've used. I feel very stupid asking these problems
 October 6th, 2007, 04:39 AM #2 Senior Member   Joined: Dec 2006 Posts: 1,111 Thanks: 0 The way most students solve these problems is to memorize all the equations relating to velocity, acceleration, and positions. However, the easy way to do this (and you don't even have to remember any equations if you do it this way), is to use integration, since the equations are all related to each other. This is the way I did it when I studied physics, and it worked great. Use a differential equation: Acceleration = change in velocity over a change in time: dv/dt = -9.81 m/s² dv = ∫ -9.81 dt V = -9.81(t) + C If you have initial vales of V and t, you can easily solve for 'C' (I'll do it for your second problem): Initial velocity and time: V_o = -10.4 m/sec t_o = 0 sec -10.4 = -9.81(0) + C C = -10.4 So, now we have: V = -9.81(t) + -10.4 Now, you have the equation to solve your second problem, since inputting any time into this equation will give you the bomb's velocity at that time. However, in the first problem, we don't want to know velocity. We want to know positions, such as how tall the building is. In that case, remember that: Velocity = change in position divided by change in time. V = dx/dt Velocity equation: V = -9.81(t) + C The velocity equation is the one we got above. Can you solve for the 'C' in problem #1, and then replace V with dx/dt and integrate to find 'x'? Remember to find the second value of 'C' resulting from the integration! Now see if you can integrate again, solve for 'C' and then solve the second problem. If you get used to solving problems this way, then when you hit diferential equation, you will have no problems, and you will be ahead of the game!
 October 28th, 2007, 08:21 PM #3 Member   Joined: Oct 2007 From: CA, USA Posts: 46 Thanks: 0 I couldn't really follow what you were talking about infinity. =[ Anyways Johnny, there are four equations that are basic for physics: V = Vo + at [what my teacher calls v-vat] V^2 = Vo^2 + 2ax [what my teacher calls very useful] x = Vot + .5at^2 [what my teacher calls very useless] avg. v = (V + Vo)/2 For number 1, you are given a, t, and Vo, so you should use very useless unless you want to solve for final velocity and then use v-vat x = Vot + .5at^2 x = 0(4.2s) + .5 (9.8m/s^2)(4.2s^2) x = 4.9m/s^2 (17.64s^2) x = 86.436 m For number 2, you have a, Vo, and t, so you should use v-vat V = Vo + at V = 10.4m/s + 9.8m/s^2(40.3s) V = 10.4m/s + 394.94m/s V = 405.34 m/s I hope that helps and that you didn't turn in that homework yet, but you probably did by now... hehe
October 29th, 2007, 03:25 AM   #4
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What you are failing to realize is that all those equation you say are useful are really just integrals of each other, so you can just start from a = -9.81 m/sec^2 and integrate to wherever you want to go, and that much more easily than you can remember all those formulas and plug them into each other. The way you did it does work, but it's the hard way. That's the way I did it before I learned basic differential equations, which is the method I was using, and it greatly simplifies the process. Try it; I think you'll like it.

 October 29th, 2007, 10:24 AM #5 Global Moderator   Joined: Dec 2006 Posts: 19,713 Thanks: 1806 [quote="mahuirong"]avg. v = (V + Vo)/2[quoted] Hence fifth equation: x = ((V + Vo)/2)t = (avg. v)t. The equations are usually taught before calculus.

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