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

The equations I have is final velocity squared = initial velocity squared plus 2(acceleration) * displacement. I can't get the equation to work with the information I'm given.

I tried setting (final velocity of the first)^2 - (initial velocity of the first)^2 / 2 times the acceleration = (final velocity of the second)^2 - (initial velocity of the second)^2 / two times the acceleration. the acceleration part canceled out, and I was left with just the final velocity of the first^2 minus the initial velocity of the first^2 (625) equals the final velocity of the second^2 which led nowhere.

Write equations for height above the ground vs. time for each ball, and set them equal. Solve for the only variable, time (t).

For the ball thrown upwards,

Y1 = 25 t - (g/2) t^2 = 25 t - 4.9 t^2

For the other ball,
Y2 = 15 - gt

Take it from there

hmm.. I got 4.0 s... if you divide each side by t, then you'd get 25-4.9t=15-9.81, then you could just subtract and divide from there?

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1 answer

  1. Daniel

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