A dart gun is fired while being held horizontally at a height of 0.85 m above ground level, and at rest relative to the ground. Thedart from the gun travels a horizontal distance of 4.27 m. A...

A dart gun is fired while being held horizontally at a height of 0.85 m above ground

level, and at rest relative to the ground. The
dart from the gun travels a horizontal distance of 4.27 m. A child holds the same gun in a horizontal position while sliding down a
42.8◦ incline at a constant speed of 3.09 m/s.

What horizontal distance x will the dart
travel if the child fires the gun forward when
it is 0.819 m above the ground?

The acceleration due to gravity is 9.8m/s2

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valentin68 | College Teacher | (Level 3) Associate Educator

Posted on

First the dart gun is fired horizontally at rest, at a height H1 =0.85 m above the ground. The trajectory of the dart is half of a parabola.

On the y axis the motion is free fall.

`H1 = (g*(t1)^2)/2` , hence `t1 =sqrt((2*H1)/g) =sqrt(2*0.85/9.81) =0.1733 s`

On the x axis the motion is uniform

`X1 = V_(1x)*t` , hence `V_(1x) =(X1)/t =4.27/0.1733 =24.64 m/s`

Now the child slides down on the inclined plane. The intial x and y components of the child speed are

`V_(0x) =V*cos(alpha) =3.09*cos(42.8) =2.267 m/s`

`V_(0y) =V*sin(alpha) =3.09*sin(42.8) =2.099 m/s`

On the y axis there will be again free fall from height H2= 0.819 m but with initial speed `V_(0y)`

`H2 =V_(0y)*t2 +(g*t2^2)/2`

`0.819 =2.099*t+9.81*(t2^2)/2`

`4.905*t2^2 +2.099*t -0.819 =0`

The roots are `t=-0.675 s` and `t=0.2473 s`

Therefore `t2 =0.2473 s`

Now, on the x axis the motion is again uniform but the initial speed is

`V_(2x)=V_(1x) +V_0x = 24.64+2.267 =26.907 m/s`

The horizontal distance of the dart in this case will be

`X2 = V_(2x)*t2 =26.907*0.2473 =6.654 m`

Sources:

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