A ball of mass 100 gm is projected vertically upwards from the ground with a velocity of `49m//sec`. At the same time another identical ball is dropped from a height of 98 m to fall freely along the same path as that followed by the first ball. After some time the two balls collide and stick together and finally fall to the ground. Find the time of flight of the masses.

Answer

6.53

Explanation

The mass of both balls are the same. Hence initially the centre of mass will be at middle of height, i.e., 49 m from ground.
The acceleration of both the ball is ‘g’ hence the acceleration of their centre of mass is also g downwards. Initial velocity of centre of mass.
`u_(c) = (m_Av_A + m_Bv_B)/(m_A + m_B) " "u_c = (m xx 0 + m xx 49)/(m + m) = (49)/2 = 24.5 m//s`.
Now considering the motion of centre of mass of balls.
Let centre of mass strikes after time t with the ground.
Using `s_y = u_y t + 1/2 a_y t^2`
`-49 = 24.5t - 4.9 t^2 " "or " "t^2 - 5t - 10 = 0`
`t = (5 +- sqrt(25 + 40))/(2)`
The positive value of t from this equation comes out to be 6.53 s. Therefore, time of flight of the ball is 6.53 s.