Kinematics (rectilnear motion) simple

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SUMMARY

The discussion focuses on determining the initial velocity v2 of a second ball launched three seconds after the first ball, which has an initial vertical velocity of v1=160 ft/sec, in order for both balls to collide at an altitude of 300 ft. The equation used for calculating the position of the first ball is Δx = (1/2)at² + v₀t, where acceleration a is set to -g (with g=32.17 ft/s² for feet). The participants emphasize the importance of calculating the time t for the first ball to reach 300 ft and adjusting for the launch delay of the second ball.

PREREQUISITES
  • Understanding of kinematic equations, specifically Δx = (1/2)at² + v₀t
  • Knowledge of gravitational acceleration values (g=32.17 ft/s² or g=9.8 m/s²)
  • Familiarity with vertical motion concepts in physics
  • Basic algebra skills for solving equations
NEXT STEPS
  • Calculate the time taken for the first ball to reach 300 ft using the kinematic equation.
  • Determine the necessary initial velocity v2 for the second ball to collide with the first at 300 ft.
  • Explore the effects of varying initial velocities on the collision height.
  • Study projectile motion principles to understand the relationship between time, velocity, and height.
USEFUL FOR

Students studying physics, educators teaching kinematics, and anyone interested in understanding the principles of rectilinear motion and collision dynamics.

whynot314
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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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