An object was launched with a velocity of 20 ms−1 at an angle of 45° to the vertical. At the
top of its trajectory the object broke into two equal pieces. One piece fell vertically
downwards. Where would the other piece fall? (Take g = 10 ms−2)
At the top of the trajectroy the object was travelling horizontally, with velocity 20/sqrt2 = 14.14 m/s
Because conservation of momentum must apply after it breaks up, the second piece must have horizontal velocity twice as large as before breakup, or 28.28 m/s
Use that and the time to fall back to earth to compute the horizontal distance it travels after breakup. Before breakup, it travelled half as far. Add them for the total displacement.
I am leaving the rest up to you because I refuse to use 10 m/s^2 for g. If a problem is worth doing, it is worth doing with the right numbers.