An airplane is dropping bales of hay to cattle stranded in a blizzard on the Great Plains. The pilot releases the bales at 150 m above the level ground when the plane is flying at 75 m/s in a direction 55° above the horizontal. How far in front of the cattle should the pilot release the hay so that the bales land at the point where the cattle are stranded?

Respuesta :

Answer:

Δx=629.35 m

The pilot release the hay 629.35 m in front of the cattle so that the bales land at the point where the cattle are stranded.

Explanation:

Step 1:

Finding initial velocity components:

Initial velocity=v=75 m/s

α=55

[tex]v_{ox}=vcos\alpha\\v_{ox}=75cos55^o\\v_{ox}=43.018 m/s\\v_{oy}=vsin\alpha\\v_{oy}=75sin55^o\\v_{oy}=61.436 m/s[/tex]

Step 2:

[tex]y_o=150\ m[/tex]

Newton Second Equation:

[tex]y-y_o=v_{oy}t+\frac{1}{2}g t^2[/tex]

g=-9.8 m/s^2 (Downward direction)

[tex]v_{oy}=61.436\ m/s[/tex]

y=0 m

Above equation will become:

-150=(61.436)t-(4.90)t^2

Solving the above quadratic equation we will get:

t=-2.09 sec           ,        t=14.63 sec

t= 14.63 sec

Step 3:

Finding the distance:

Using Again Newton equation of motion in x-direction:

[tex]x-x_o=v_{ox}t+\frac{1}{2}a_{x} t^2[/tex]

Since velocity is constant in x- direction, [tex]a_x[/tex] will be zero.

Above equation will be:

[tex]\Delta x=v_{ox}t[/tex]

Δx=(43.018)(14.63)

Δx=629.35 m

The pilot release the hay 629.35 m in front of the cattle so that the bales land at the point where the cattle are stranded.

The pilot should release the hay at a height of 629.35 m.

Given information,

Initial velocity = 75 m/s

Velocity For x-component,

[tex]\bold {V_0x = Vcos \alpha}\\\\\bold {V_0x = 75 cos 55^o}\\\\\bold {V_0x = 43. 018m/s}[/tex]

Velocity for Y-component

[tex]\bold {V_0y = Vsin \alpha}\\\\\bold {V_0y = 75 sin 55^o}\\\\\bold {V_0y = 61. 43m/s}[/tex]

Using Newton's second equation for y-axis,

[tex]\bold {y-y_0 = V_0t + \dfrac {1}{2} gt^2}[/tex]

Where,

g - gravitational acceleration

put the values in the equation,

[tex]\bold {-150=(61.436)t-(4.90)t^2}[/tex]

Solving this quadratic equation, we get 2 values

t = 14.29 s

To find the distance, use Newton's second equation,

[tex]\bold {x-x_0 = V_0t + \dfrac {1}{2} gt^2}[/tex]

Since acceleration is zero because the velocity is constant in x-axis hence .

So,

[tex]\bold {x-x_0 = V_0_xt }[/tex]

[tex]\bold {x- x_0=(43.018)(14.63)}\\\\\bold {x - x_0=629.35 m}[/tex]

Therefore, the pilot should release the hay at 629.35 m.

To know more about Newton's law, refer to the link:

https://brainly.com/question/3715235