from the top of a tower of height 78. 4 metres to stones are projected horizontally with speed 10 metre per second and 15 M per second in opposite directions the distance of separation between them on reaching the ground is

Dear student,

Height of the tower is 78.4 m.
the only acceleration is in the vertical direction which is g. Using eqn of motions

Height(78.4)= ut+1/2(gt²)

Where u is intial vertical component of velocity. Since the stones are projected horizontally it is zero.

78.4= 1/2(9.8×t²)
t²= 16
t= 4 s.

Now the first stone will move a distance:

R¹(range)=u(x)×t
Where u(x) is horizontal initial velocity.

R¹= -10×4= -40m.(we took negative assuming the first stone is thrown to left of tower, which we have taken as negative direction.)

R²= 15×4= 60m.

Distance of separation=
60-(-40)= 100m.

A better idea can be obtained if you draw a diagram and try to solve yourself.

Thank you.