A blue ball is thrown upward with an initial speed of 22.8 m/s, from a height of

Question

A blue ball is thrown upward with an initial speed of 22.8 m/s, from a height of 0.9 meters above the ground. 2.8 seconds after the blue ball is thrown, a red ball is thrown down with an initial speed of 12.2 m/s from a height of 29.1 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. What is the maximum height the blue ball reaches?What is the height of the red ball 3.64 seconds after the blue ball is thrown? How long after the blue ball is thrown are the two balls in the air at the same height?

Explanation / Answer

For the blue ball: to=0s, yo=0.6m, vo=21.6m/s For the red ball: to=2.6s, yo=26.4m, vo=8.3m/s The motion of both balls are given by: y = yo+vo*t-(g/2)t² To solve for time t, set heights equal: 0.6+21.6t-(g/2)t² = 26.4-8.3(t-2.6)-(g/2)(t-2.6)² After multiplying out and collecting terms, we get: t = 3.2 s (y = 19.1m)

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