How Long Until Two Balls Thrown Differently Meet at the Same Height?

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SUMMARY

The problem involves two balls: one thrown upward from the ground with an initial speed of 25 m/s and another dropped from a height of 15 m. To find when they meet at the same height, kinematic equations are applied. The first ball's height is given by the equation y(t) = 25t - 0.5(9.81)t², while the second ball's height is y(t) = 15 - 0.5(9.81)t². Setting these equations equal allows for solving the time variable t, leading to the solution of when the two balls meet.

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Homework Statement


ball is thrown upward from the ground with an initial speed of 25m/s; at the same instant a ball is dropped from rest from a building 15m high. After how long will the balls be at the same height?


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The Attempt at a Solution


well the first ball has initial(vi) of 25m/s and if its thrown that at its peak it wud be 0m/s and the acceleration is -9.81 and i used vf=vi+at which will give me 2.55
and for the second ball the initial velocity is 0 so i used x=1/2a(t)^2 and i got 1.75

i have no clue like what to do after you get both the times like how do u find where they would meet up??

thx for hte help
 
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So you have two [kinematic] equations with two unknowns, how would you normally solve a system of simultaneous equations?
 
give an expression for the heigth of the ball as a function of time and then set them equal, and then solve for t.

in general you have y(t) = y(0) + v(0)t - 1/2gt^2 for an object released at time t=0 at height y(0) with initial speed v(0). you know y(0) and b(0) for both balls.
 

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