Applied mechani?

Let's denote the initial velocity of the projected body as u and the time taken for the two bodies to meet as t.

For the body dropped from the top of the tower:
Initial velocity, u1 = 0 m/s
Acceleration, a1 = 9.81 m/s^2 (acceleration due to gravity)
Distance fallen, s1 = 100 m

Using the equation of motion: s1 = u1*t + (1/2)*a1*t^2
100 = 0*t + (1/2)*9.81*t^2
100 = 4.905*t^2
t = sqrt(100/4.905) = 4.52 seconds

For the projected body:
Initial velocity, u
Acceleration, a2 = -9.81 m/s^2 (negative because it is moving upwards)
Distance covered, s2 = 50 m (half-way point)

Using the equation of motion: s2 = u*t + (1/2)*a2*t^2
50 = u*4.52 + (1/2)*(-9.81)*4.52^2
50 = 4.52u - 99.5
4.52u = 149.5
u = 33.09 m/s

Therefore, the initial velocity of the projected body is 33.09 m/s.

To find the velocity when the two bodies meet, we can use the equation:
v = u + at

For the projected body:
v = 33.09 - 9.81*4.52
v = -11.97 m/s

Therefore, the velocity of the projected body when it meets the descending body is -11.97 m/s.