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December 22, 2014

December 22, 2014

Posted by **a** on Tuesday, September 13, 2011 at 12:22am.

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?

- phy -
**Henry**, Wednesday, September 14, 2011 at 8:14pm1. 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.

- phy -
**a**, Wednesday, September 14, 2011 at 10:25pmplease answer the 5th part

- phy -
**Henry**, Friday, September 16, 2011 at 11:52am5. 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.

Check:

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).

- phy -
**Aziz**, Wednesday, September 28, 2011 at 11:25pmI have noticed that the gravity for the blue ball is negative, while the red ball is positive? could you tell me why?

- phy -
**Jack**, Friday, January 13, 2012 at 9:19pmregarding #5:

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.

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