A helicopter is flying horizontally at 8.15 m/s and an altitude of 18.5 m when a package of emergency medical supplies is ejected horizontally backward with a speed of 10.4 m/s relative to the helicopter. Ignoring air resistance, what is the horizontal distance between the package and the helicopter when the package hits the ground?

Question

contestada

Respuesta :

Khoso123 Khoso123 · Answer 1 · 2020-02-15T05:16:33+01:00

Answer:

[tex]d=39.27m[/tex]

Explanation:

Let +y is upward direction and backward be +x direction

then we have given data

[tex]v_{hg}=-8.15m/s\\v_{ph}=10.4m/s\\\\then\\v_{pg}=v_{hg}+v_{ph}\\v_{pg}=-8.15m/s+10.4m/s\\v_{pg}=2.25m/s[/tex]

From equation simple motion we find the time

So

[tex]y_{f}=y_{i}+v_{yi}t+(1/2)a_{y} t^{2} \\ 0=18.5m+0+(1/2)(-9.8m/s^{2} )t^{2}\\-18.5=-4.9t^{2}\\t=3.7755s[/tex]

To get the distance between the point the package ejected to the point it the ground.

[tex]v_{x}=x_{p}/t\\ x_{p}=v_{x}*t\\x_{p}=(2.25m/s)*(3.7755s)\\x_{p}=8.5m(backward)[/tex]

Now we have to get the distance the helicopter travels during same interval

[tex]x_{h}=v_{hg}*t\\x_{h}=-8.15m/s*3.7755s\\x_{h}=30.77m(forward)[/tex]

Now to get the distance between the package and helicopter when package hit the ground

[tex]d=x_{p}-x_{h}\\d=8.5m-(-30.77m)\\d=39.27m[/tex]