A blue ball is thrown upward with an initial speed of 21.2 m/s, from a height of

Question

A blue ball is thrown upward with an initial speed of 21.2 m/s, from a height of 0.9 meters above the ground. 2.6 seconds after the blue ball is thrown, a red ball is thrown down with an initial speed of 9.4 m/s from a height of 24.9 meters above the ground. The force of gravity due to the earth results in the balls each having a constant downward acceleration of 9.81 m/s2. Take up as the positive direction.

How long after the blue ball is thrown are the two balls in the air at the same height?

I keep try to solve it but i keep gettie .78 which does assume they both left at the same time which they didn't. please explain your answer.

Explanation / Answer

y blue = 0.9 + 21.2*t - 0.5*9.81*t^2

y red = 24.9 - 9.4*(t-2.6) - 0.5*9.81*(t-2.6)^2

yblue = yred

0.9 + 21.2*t - 0.5*9.81*t^2=24.9 - 9.4*(t-2.6) - 0.5*9.81*(t-2.6)^2

solve for t

t=3.0 s

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