Question

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

Answer 1

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.