Answer:
a
![D = 1162.7 \ m](https://img.qammunity.org/2021/formulas/physics/college/tjp5hudbokz4978a9pgwiwqfkmcj0bhmpb.png)
b
![\beta =- 65.55^o](https://img.qammunity.org/2021/formulas/physics/college/56t3q62nbcebcgopi3lztofa55pxsiwh34.png)
Step-by-step explanation:
From the question we are told that
The speed of the airplane is
![u = 92.3 \ m/s](https://img.qammunity.org/2021/formulas/physics/college/nelx2at31df3tayjj2j4ggtbxazbnvjrug.png)
The angle is
![\theta = 51.1^o](https://img.qammunity.org/2021/formulas/physics/college/k2loji0q5wox3w04zr3qa1z13a04h3x8vu.png)
The altitude of the plane is
![d = 532 \ m](https://img.qammunity.org/2021/formulas/physics/college/6bqlkq989vxxeqyr8ysn4o34wkl45a9imy.png)
Generally the y-component of the airplanes velocity is
![u_y = v * sin (\theta )](https://img.qammunity.org/2021/formulas/physics/college/2uislle6eja3qw8aucqz6uqci2dr250mhg.png)
=>
![u_y = 92.3 * sin ( 51.1 )](https://img.qammunity.org/2021/formulas/physics/college/pzn62v6cvkmjwvekaca6owla0tptz4gzn9.png)
=>
Generally the displacement traveled by the package in the vertical direction is
![d = (u_y)t + (1)/(2)(-g)t^2](https://img.qammunity.org/2021/formulas/physics/college/rn0tnghdt8qxqjcz2jwtz2z6q78o9wt3eo.png)
=>
![-532 = 71.83 t + (1)/(2)(-9.8)t^2](https://img.qammunity.org/2021/formulas/physics/college/o7lu7ojem6ckqf9c6rrfs6bpwuxbtl34vr.png)
Here the negative sign for the distance show that the direction is along the negative y-axis
=>
![4.9t^2 - 71.83t - 532 = 0](https://img.qammunity.org/2021/formulas/physics/college/r4zw0t2e4xpo7tgjecb6uf1gibuwz1i58v.png)
Solving this using quadratic formula we obtain that
![t = 20.06 \ s](https://img.qammunity.org/2021/formulas/physics/college/nofdq9zrhjdyh31u5inxx9t6lb0xo2f2zm.png)
Generally the x-component of the velocity is
![u_x = u * cos (\theta)](https://img.qammunity.org/2021/formulas/physics/college/tv7z9zre3b4l95pohrrows9syukqqpqkn4.png)
=>
![u_x = 92.3 * cos (51.1)](https://img.qammunity.org/2021/formulas/physics/college/mu85r9isfqcmk0gge6wwjsz7qpy5ocgr6v.png)
=>
Generally the distance travel in the horizontal direction is
![D = u_x * t](https://img.qammunity.org/2021/formulas/physics/college/dlza7st4h6a6jq1rke3303zgf287aktczy.png)
=>
![D = 57.96 * 20.06](https://img.qammunity.org/2021/formulas/physics/college/vg7izwnydxoq6ccveq24qxw3a5sfsydxb5.png)
=>
![D = 1162.7 \ m](https://img.qammunity.org/2021/formulas/physics/college/tjp5hudbokz4978a9pgwiwqfkmcj0bhmpb.png)
Generally the angle of the velocity vector relative to the ground is mathematically represented as
![\beta = tan ^(-1)[(v_y)/(v_x ) ]](https://img.qammunity.org/2021/formulas/physics/college/yel0yte7bab2tizxzb7f05ufx3s003bka9.png)
Here
is the final velocity of the package along the vertical axis and this is mathematically represented as
![v_y = u_y - gt](https://img.qammunity.org/2021/formulas/mathematics/high-school/7s5y4eqnmw5qy4e0ewo3rf5rrolsiywxib.png)
=>
=>
and v_x is the final velocity of the package which is equivalent to the initial velocity
![u_x](https://img.qammunity.org/2021/formulas/physics/college/ockpfvgjnw4uv4925qrndi6ut82tb897pz.png)
So
![\beta = tan ^(-1)[-130.05}{57.96 } ]](https://img.qammunity.org/2021/formulas/physics/college/tbkz62l8bwju30tmk9oc48zh83gw0xl6nc.png)
![\beta =- 65.55^o](https://img.qammunity.org/2021/formulas/physics/college/56t3q62nbcebcgopi3lztofa55pxsiwh34.png)
The negative direction show that it is moving towards the south east direction