A classical mechanics problem by Jilvan Júnior
Classical Mechanics
Level
3
A stone is dropped from a height of 19.6 m, above the ground while a second stone is simultaneously projected from the ground with sufficient velocity to enable it to ascend 19.6 m. Where the stones would meet?
The answer is 14.7.
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1 solution
Let the ball on the top be A and the ball on the bottom be B Now let us assume that they meet at a distance of "x" meters from the top So the remaining distance is 19.6 - x mtrs Now for ball B , it is said that it has sufficient velocity to reach the top Therefore by applying v^2 = u^2 - 2as where v = 0 , a = 9.8m/s^2 and s = 19.6 we get u to be 19.6m/s Now for ball A, to cover a distance of x meters the time required by it is sqrt(x/4.9) . (by applying s = ut + .5 at^2 so for ball B the time is the same as both the balls meet at that position So by replacing the value of "t" by sqrt(x/4.9) along with 'u', 'a' and 's' in the equation s= ut +.5at^2 we get the value of x to be 4.9 Hence the distance from the bottom is 19.6-4.9 = 14.7mts