A red ball is dropped from rest at a height of 6.20 m.A blue ball at a height of

Question

A red ball is dropped from rest at a height of 6.20 m.A blue ball at a height of 11.1 m is thrown down at the same instant at 8.00 m/s. How long does it take the blue ball to catch up with the red ba Need Help? moves with high above this point is her center of mass at the following times? (ignore the effects of air resistance, and assume the initial height of her center of mass isat y-o) (a) 0.100 s differs from the correct answer by more than 10%. Double check your calculations. m (b) 0.200 () 0.300 s (d) 0.500 s

Explanation / Answer

Acceleration due to gravity when moving upward should be negative.

So, g = -9.81 m/s^2

V_initial = 4.61 m/s

So using s=at - 0.5gt2

1) at 0.100 s

D = 4.61*.1 - 4.905*(.100)^2 = 0.41195 ~ 0.412 m

2)

D = 4.61*.2 - 4.905*(.200)^2 = 0.7258 ~ 0.726 m

3)

D = 4.61*.3 - 4.905*(.300)^2 = 0.94155 ~ 0.942 m.

4)

D = 4.61*.5 - 4.905*(.500)^2 = 1.07875 ~ 1.079 m.

The other problem:

The red ball started from rest, hence, we can write,

6.20 = 1/2*9.8*t2

Or, 6.20 -1/2*9.8*t2 = 0;

The Blue ball was thrown with a velocity of 8m/s, hence we can write,

11.1 = 8t + 1/2*9.8*t2

Or, 11.1 - 8t - 1/2*9.8*t2 = 0;

Thus,

6.20 -1/2*9.8*t2 = 11.1 - 8t - 1/2*9.8*t2

we solve for 't' ,

where t = 0.6125 seconds.

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