A dart gun is fired while being held horizontally at a height of 1.00m above ground level, and at rest relative to the ground.The dart from the gun...

First part:

The dart moves
horizontally 5 m from a height of 1 m.

So it took took t seconds to
fall by 1 m. Therefore the time taken to reach the ground is given by height h = (1/2)gt^2. Or t
= sqrt(2h/g). = sqrt(2/g), g is acceleration due to gravity = 9.81
m/s^2.

 h = 1m =(1/2)gt^2.

Therefore
the time t taken by the dart to reach the ground is given by  t = sqrt(2/g)
seconds.

Since the horizontal speed is not affected by gravity, it
is constant. So the to cover the distance of  5 meter in sqrt(2/g) secs, the constant horizontal
speed of the dart must be  distance/time =  5/sqrt2/g) =
5(sqrt(g/2).

So the speed of the dart horizontally = 
5sqrt(g/2).

We use this speed  x in the  second
part.

Second part:

The child slide
horizontally at a constant speed of 2 m/s.

So the the horizontantal
and vertical speed  component of the child is imparted to the dart  from the gun in addition to
the speed x of the the dart projected horizontally.

The horizontal
and vertical components of the child is 2*cos45 and 2 sin45 = 2/sqr2 and 2/sqrt2, each equal to
sqrt2.

Let x be the initial speed of dart projected horizontally. So
the dart's vertical component must be zero.

The child's horizontal
and vertical speed components must be added to that of dart.

So the
total initial horizontal and vertical speed components of the dart= sqrt2 +x and sqrt2 +0 =
sqrt2. respectively.

Therefore the the time the  the dart takes to
fall from 1 meter with an initial speed of u = sqrt2 m/s to fall to ground is given
by:

ut^2+(1/2) gt^2 = 1meter.

ut+(1/2)
g t^2 = 1.

gt^2 +2ut-2 = 0.

t =
{-2u+sqrt(4u^2+8g)}/2g by quadratic formula.

 t =
{-u+sqrtu^2+2g)}/g. But u = sqrt2.

t =
{-sqrt2+sqrt(2+2g)}/g}.

Therefore,
the dart has a  constant horizontal speed of sqrt2 imparted from the child +  speed imparted by
the gun x  =  sqrt2 + 5sqrt(g/2)  for a period of  time t =
{-sqrt2+sqrt(2+2g)}/2g.

So the horizontal distance travelled by the
dart =  speed *time = {(sqrt2+5sqrt(g/2)}{-sqrt2+sqrt(2+2g)}/g

=
{12.4878)(0.3298) =  4.12 m. So it is less than 5m because , the time of falling vertically is
reduced by the increased initial vertical component of the speed  contribution imparted by the
the child's  sliding speed component to the extent of sqrt 2.

The
horizontal distance traversed by the dar = 4.12 m.