1. The problem statement, all variables and given/known data 2. Relevant equations I'm not sure what equations to use here, or even wether or not I need to use calc. Unfortunately we just started dynamics and most students don't know how to integrate yet (including me) v = dx/dt a = dv/dt = d^{2}x/dt^{2} = v dv/dx Uniform rectilinear motion: x = x0 + vt Uniformly accelerated rectilinear motion: v = vo + at ; x = xo + vot + 1/2at^{2}; v^{2} = v^{2}o +2a(x-xo) where o in all the above = initial. 3. The attempt at a solution I don't even know where to start with this one. hence my question for help. We know that a = 9.81m/s^{2} and also that every .5 second there is a ball dropped. Do i have to integrate anything?
As you have stated already, a is a constant (9.81 m/s^2). With uniform acceleration, there is no need to know how to integrate.
ok. Well I think I have a solution then, but it's not very fancy (I just used techniques from high school physics). 1) I first contemplate ball #1 to find the final velocity and the time it is at 3m. V_{f}^{2}=V_{i}^{2}+2ax V_{f}^{2}= 0^{2}+2(9.81)(3) V_{f}=7.67m/s Now the time: x=( (V_{f}+V_{i})/2)t ; 3m=((7.67+0)/2)t ; t= .78206s Now I say Ball #2 is dropped .5s after ball #1 and solve for it's final velocity a=(V_{f}-V_{i})/t ; V_{f}= 2.767m/s Now I solve for displacement between ball 1 and 2 according to their respective instantanous(or final) velocities and time being dropped apart from each other. Where x is displacement. x=( (V_{f}+V_{i})/2)t x= ((7.67+2.767)/2)*0.5 x= 2.6097m Can someone confirm this is correct methodology and answer? Thanks,
High school physics is all I use everyday This part is correct. (It was probably easier to use the formula for displacement, x = V_{i} t + 1/2 a t^{2}.) I don't quite understand your explanation for this part of your solution, but if you're saying what I think you're saying, the answer is correct! Are you saying that the problem is equivalent to the distance moved by a ball starting with an initial velocity of 2.767m/s in 0.5 sec? A simpler solution might be to just find what distance ball #2 travels in 0.28206 seconds.
Actually.. i didn't articulate that very well i suppose. But you're right, if I just solved for the distance ball #2 travelled and deducted that figure from 3m it should tell me the spread between them and hence h (or x).