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A blue ball is thrown upward with an initial speed of 20.6 m/s, from a height of 0.8 meters above the ground. 2.5 seconds after the blue ball is thrown, a red ball is thrown down with an initial speed of 8.2 m/s from a height of 23.7 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
1)What is the speed of the blue ball when it reaches its maximum height?
2)How long does it take the blue ball to reach its maximum height?
3)What is the maximum height the blue ball reaches?
4)What is the height of the red ball 3.25 seconds after the blue ball is thrown?
5)How long after the blue ball is thrown are the two balls in the air at the same height?
1. Final velocity(Vf) = 0 @ max. ht.
2. t = (Vf - Vo) / g,
t = (0 - 20.6) / -9.8 = 2.1s.
3. h = Vo*t + 0.5gt^2,
h = 20.6*2.1 + 0.5*(-9.8)*(2.1)^2,
h = 43.26 + (-21.6) = 21.65m above 0.8m.
4. t = 3.25 - 2.5 = 0.75s.
h = 23.7 - (V0*t + 0.5gt^2),
h=23.7 - (8.2*0.75 + 0.5*9.8*(0.75)^2),
h = 23.7 - (6.15 + 2.76),
h = 23.7 - 8.91 = 14.79m Above ground.
please answer the 5th part
5. h1 + h2 = 23.7m,
(0.8+20.6t-4.9t^2)+(8.2t+4.9t^2)=23.7, 28.8t = 23.7 -0.8 = 22.9,
t = 0.795s.
h1=0.8 + 20.6*0.795 -4.9*(0.795)^2 = 14.1m above ground(t = 0.795s).
h2=23.7 - 8.2*0.795 + 4.9*(0.795^2 =
23.7 -9.62=14.1m above groun(t=0.795s).
I have noticed that the gravity for the blue ball is negative, while the red ball is positive? could you tell me why?
That answer would be correct if both balls were thrown at the same time. The problem states the red ball was thrown downward 2.6 seconds after the blue ball was thrown. One solution is to determine the position and velocity of the blue ball at the time the red ball was thrown, and then use those as the initial condition.