Question

Paintball guns were originally developed to mark trees for logging. A forester aims his gun directly at a knothole in a tree that is 4.0 m above the gun. The base of the tree is 15 m away. The speed of the paintball as it leaves the gun is 50 m/s. How far below the knothole does the paintball strike the tree? Express your answer with the appropriate units.

Answer 1

The distance between knothole and the paint ball is 0.483 m.

Step-by-step explanation:

Given that,

Height = 4.0 m

Distance = 15 m

Speed = 50 m/s

The angle at which the forester aims his gun are,

`\tan\theta=(4)/(15)`

`\tan\theta=0.266`

`\cos\theta=(15)/(√(15^2+4^2))`

`\cos\theta=0.966`

Using the equation of motion of the trajectory

The horizontal displacement of the paint ball is

`x=(u\cos\theta)t`

`t=(x)/(u\cos\theta)`

Using the equation of motion of the trajectory

The vertical displacement of the paint ball is

`y=u\sin\theta(t)-(1)/(2)gt^2`

`y=u\sin\theta((x)/(u\cos\theta))-(1)/(2)g((x)/(u\cos\theta))^2`

`y=x\tan\theta-(gx^3)/(2u^2(\cos\theta)^2)`

Put the value into the formula

`y=(15*0.266)-((9.8*(15)^2)/(2*(50)^2*(0.966)^2))`

`y=3.517\ m`

We need to calculate the distance between knothole and the paint ball

`d=h-y`

`d=4-3.517`

`d=0.483\ m`

Hence, The distance between knothole and the paint ball is 0.483 m.