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Registered Member #2431
Joined: Tue Oct 13 2009, 09:47PM
Location: Chico, CA. USA
Posts: 5639
Im having a difficult time with what should be a simple problem. My brain is just to sore to deal with 1D kinematics. So, coming here to the 4hv.org forum where other smart people are, i decided to simplify the problem.
Now, in a simplified problem: Throwing a 750g ball straight up with 25 N-sec of force with the force applied in 1.6 seconds. how high will it go ? (ignore aerodynamics)
Im trying to think in terms of the v(t), x(t), a(t) graphs.
Registered Member #11591
Joined: Wed Mar 20 2013, 08:20PM
Location: UK
Posts: 556
N-sec isn't force. N, newtons, that's force. assuming your units were right: 25/1.6 = the force in newtons = 15.625 N force = mass * acceleration, therefore force / mass = acceleration 15.625/0.75 = 20.833 m/s^2 multiplying by the time it took to accelerate, we find the ball reached the following speed at the end of the throw: 20.833*1.6 = 33.33 m/s The easiest way to proceed is to find the energy in the ball: K.E. = 0.5mv^2 0.5*0.75*33.33^2 = 416.67 J now we also know that potential energy is equal to height multiplied by mass and g so assuming that nothing is lost: energy / (mass*g) = height: height = 416.67/(0.75*9.812) = 56.6 m
I'm sorry if this explanation isn't very general, and makes use of the energy shortcut.
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Math and Physics of Upward Acceleration.
