0

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 at^{2}

Under the influence of gravity, the equation becomes:

s = ut - 1/2 gt^{2}

g = 10m/s^{2}

Since s1 = s2,

ut - 1/2 gt^{2} = u(3 + t) - 1/2 g(3 + t)^{2}

50t - 1/2 (10)t^{2} = 50(3 + t) - 1/2 (10)(3 + t)^{2}

50t - 5t^{2} = 150 + 50t - 5(3 + t)^{2}

- 5t^{2} = 150 - 5(3 + t)^{2}

- 5t^{2} = 150 - 5(9 + 6t + t^{2})

-5t^{2} = 150 - 45 - 30t - 5t^{2}

30t = 105

t = 105/30

t = 3.5s

Just before hitting the ground, s = 0

For the first ball,

s = ut - 1/2 gt^{2}

0 = 50t - 1/2 (10)t^{2}

0 = 50t - 5t^{2}

5t^{2} = 50

t^{2} = 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 + t^{2})

0 = 150 + 50t - 45 - 30t - 5t^{2}

0 = 105 + 20t - 5t^{2}

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

Your Answer