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Ball A is thrown up vertically with velocity 50m/s and after...
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Ball A is thrown up vertically with velocity 50m/s and after 3s a second ball is thrown up calculate the time the two balls passes each other and the time difference of the balls arriving on the ground

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Ghana
04 October 2020

School not set Nigeria
04 October 2020
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Let s1 be the height of the first ball from the initial point and s2 be the height of the second ball from the initial point.

The time the two ball just passes each other, s1 = s2

Also, from second equation of motion, s = it + 1/2 at2

Under the influence of gravity, the equation becomes:

s = ut - 1/2 gt2

g = 10m/s2

Since s1 = s2,

ut - 1/2 gt2 = u(3 + t) - 1/2 g(3 + t)2

50t - 1/2 (10)t2 = 50(3 + t) - 1/2 (10)(3 + t)2

50t - 5t2 = 150 + 50t - 5(3 + t)2

- 5t2 = 150 - 5(3 + t)2

- 5t2 = 150 - 5(9 + 6t + t2)

-5t2 = 150 - 45 - 30t - 5t2

30t = 105

t = 105/30

t = 3.5s

Just before hitting the ground, s = 0

For the first ball,

s = ut - 1/2 gt2

0 = 50t - 1/2 (10)t2

0 = 50t - 5t2

5t2 = 50

t2 = 10

t = 3.162 sec

For the second ball,

0 = 50(3 + t) - 1/2 (10)(3 + t)2

0 = 150 + 50t - 5(9 + 6t + t2)

0 = 150 + 50t - 45 - 30t - 5t2

0 = 105 + 20t - 5t2

t = -3 or t = 7

But time cannot be negative, thus, - 3 is ignored

t = 7

Therefore, the time difference is:

7 - 3.162 = 3.838 sec

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