A red ball is thrown down with an initial speed of 1.0 m/s from a height of 26.0

Question

A red ball is thrown down with an initial speed of 1.0 m/s from a height of 26.0 meters above the ground. Then, 0.6 seconds after the red ball is thrown, a blue ball is thrown upward with an initial speed of 23.5 m/s, from a height of 1.0 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/s^2.

1) What is the speed of the red ball right before it hits the ground? __________________m/s

2) How long does it take the red ball to reach the ground? __________________s

3) What is the maximum height the blue ball reaches? __________________m

4) What is the height of the blue ball 1.9 seconds after the red ball is thrown? __________________m

5) How long after the red ball is thrown are the two balls in the air at the same height? __________________s

Explanation / Answer

1. V^2 = Vo^2 + 2a*d. V^2 = 1^2 + 19.62*27 = 530.74 V = 23 m/s. 2. V = Vo + gt. t = (V-Vo)/g. t = (23-1) / 9.81 = 2.24 s. 3. hmax = ho + (V^2-Vo^2)/2g. hmax=0.7 + (0-(24.5)^2)/-19.62= 31.3 m. 4. h = ho + Vo*t + 0.5g*t^2. h=0.7+ 24.5*(2-0.6) -4.9(2-0.6)^2 = 25.39.

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