A ball weighing 100 g was thrown veritcally upwards with a velocity 49 ms^(-1). At the same moment another indentical ball was dropped from a height of 98 m vertically above the first ball. After some time the two balls collided and got stuck together. This combined mass reached the ground finally. Determine how long the balls were in motion.

Explanation

Let the height attained by the balls above the ground in time `t_(1)` be h when they collide with each other. Analysing the upward motion of the first ball, we get
h = 49 `t_(1) - 9.8 t_(1)^(2)`/2
Analysing the downward motion of the second ball, we get
98 - h = 9.8`t_(1)^(2)`/2
From equations (1) and (2),
98 = 49`t_(1) " " or , t_(1) = 2`s
At the time of collision the velocity of the ball thrown in the upward direction is `v_(1)` and that of the ball thrown in the downward direction is `v_(2)`.
`therefore" "v_(1) = 49 - 9.8 xx 2 = 29.4 "m"cdot s^(-1)` (upward)
and `v_(2) = 9.8 xx 2 = 19.6 "m"cdot s^(-1) ("downward")`
After collision, the velocity of the combined mass is V . Then according to the law of conservation of momentum,
0.1 `xx 29 - 0.1 xx 19.6 = 2 xx 0.1 xx V`
or, V = 4.9 m`cdot s^(-1)` (upward)
From equation (1), we get
h = 49`xx 2 - 9.8 (2)^(2)//2 = 78.4` m
Let the time taken by the combined mass to reach the ground be `t_(2)`. Then
78.4 = -4.9`t_(2) + 9.8 xx t_(2)^(2)//2 " "or, t_(2)^(2)-t_(2) - 16 = 0 `
or, `t_(2) = (1 pm sqrt(1 + 64)//2 = 4.53 s " " (because t cancel(lt) 0 )`
`therefore" " "Total time" = t_(1) + t_(2) = 2 + 4.53 = 6.53`s.