Kinematics (rectilnear motion) simple

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Ball 1 is launched with an initial vertical velocity of 160 ft/sec and reaches an altitude of 300 ft after a certain time, which can be calculated using the equation Δx = (1/2)at² + v₀t. To find the initial velocity v2 of Ball 2, which is launched three seconds later, the time for Ball 1 to reach 300 ft must be determined first, then adjusted by subtracting three seconds. The acceleration due to gravity should be considered as -32.17 ft/s² when calculating the motion. The discussion emphasizes the importance of using basic kinematic equations for clarity and accuracy in solving the problem. Overall, the approach focuses on correctly applying kinematic principles to determine the conditions for the balls to collide at the specified altitude.
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Ball 1 is launched with an initial vertical velocity v1=160 ft/sec. Three seconds later, ball 2 is launched with an initial velocity v2. determine v2 if the balls are to collide at an altitude of 300ft.


not sure where I am going wrong, do I have my limits correct?[URL=http://s1341.photobucket.com/user/nebula-314/media/20131222_191532_zpsf76c1ab5.jpg.html][PLAIN]http://i1341.photobucket.com/albums/o745/nebula-314/20131222_191532_zpsf76c1ab5.jpg[/URL][/PLAIN]
 
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v2 being initial velocity.
 
I'm not sure what you're doing here, but if I were you I would use the following equation to calculate t for the first ball to reach 300ft, then using that t minus 3 (because the 2nd ball was launch 3 seconds later) and the same equation I would calculate v2.

\Deltax= \frac{1}{2}at^{2}+v_{0}t

Just remember that here we take upside direction to be positive, hence a=-g and if you are measuring distance in ft you should use g=32.17 ft/s^2 , or convert ft to meter and use g=9.8 m/s^2

Let me know how it goes :)
 
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Oh i was trying to derive that equation, and perhaps save some steps.
 
But your solution did help a lot thanks.
 
oh I see! I guess sometimes it's just easier to stick to the basics! :-p

No problem at all, happy to help
 
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