# 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 travels a horizontal distance of 5.00m. A child...

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 travels a horizontal distance of 5.00m. A child hold the same gun in a horizontal position while sliding down a 45° incline at a constant speed of 2.00 m/s. How far will the dart travel if the child fires the gun when it is 1.00 m above the ground? Please explain what you do.

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### 1 Answer

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.