Answer:
a
![D = 1162.7 \ m](https://tex.z-dn.net/?f=D%20%3D%20%201162.7%20%5C%20%20m%20)
b
![\beta =- 65.55^o](https://tex.z-dn.net/?f=%5Cbeta%20%3D-%2065.55%5Eo)
Explanation:
From the question we are told that
The speed of the airplane is ![u = 92.3 \ m/s](https://tex.z-dn.net/?f=u%20%20%3D%20%2092.3%20%5C%20m%2Fs)
The angle is ![\theta = 51.1^o](https://tex.z-dn.net/?f=%5Ctheta%20%3D%2051.1%5Eo)
The altitude of the plane is ![d = 532 \ m](https://tex.z-dn.net/?f=d%20%3D%20%20532%20%5C%20%20m)
Generally the y-component of the airplanes velocity is
![u_y = v * sin (\theta )](https://tex.z-dn.net/?f=u_y%20%20%3D%20%20v%20%2A%20%20sin%20%28%5Ctheta%20%29)
=> ![u_y = 92.3 * sin ( 51.1 )](https://tex.z-dn.net/?f=u_y%20%20%3D%20%20%2092.3%20%2A%20%20sin%20%28%2051.1%20%29)
=>
Generally the displacement traveled by the package in the vertical direction is
![d = (u_y)t + \frac{1}{2}(-g)t^2](https://tex.z-dn.net/?f=d%20%3D%20%20%28u_y%29t%20%2B%20%20%5Cfrac%7B1%7D%7B2%7D%28-g%29t%5E2)
=> ![-532 = 71.83 t + \frac{1}{2}(-9.8)t^2](https://tex.z-dn.net/?f=%20-532%20%20%3D%2071.83%20t%20%2B%20%20%5Cfrac%7B1%7D%7B2%7D%28-9.8%29t%5E2)
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://tex.z-dn.net/?f=4.9t%5E2%20-%2071.83t%20-%20532%20%3D%200)
Solving this using quadratic formula we obtain that
![t = 20.06 \ s](https://tex.z-dn.net/?f=t%20%3D%20%2020.06%20%5C%20%20s)
Generally the x-component of the velocity is
![u_x = u * cos (\theta)](https://tex.z-dn.net/?f=u_x%20%20%3D%20%20u%20%20%2A%20%20cos%20%28%5Ctheta%29)
=> ![u_x = 92.3 * cos (51.1)](https://tex.z-dn.net/?f=u_x%20%20%3D%20%20%2092.3%20%20%2A%20%20cos%20%2851.1%29)
=>
Generally the distance travel in the horizontal direction is
![D = u_x * t](https://tex.z-dn.net/?f=D%20%3D%20%20u_x%20%20%2A%20%20t)
=> ![D = 57.96 * 20.06](https://tex.z-dn.net/?f=D%20%3D%20%2057.96%20%20%2A%20%20%2020.06%20)
=> ![D = 1162.7 \ m](https://tex.z-dn.net/?f=D%20%3D%20%201162.7%20%5C%20%20m%20)
Generally the angle of the velocity vector relative to the ground is mathematically represented as
![\beta = tan ^{-1}[\frac{v_y}{v_x } ]](https://tex.z-dn.net/?f=%5Cbeta%20%20%3D%20%20tan%20%5E%7B-1%7D%5B%5Cfrac%7Bv_y%7D%7Bv_x%20%7D%20%5D)
Here
is the final velocity of the package along the vertical axis and this is mathematically represented as
![v_y = u_y - gt](https://tex.z-dn.net/?f=v_y%20%20%3D%20%20u_y%20%20-%20%20%20gt)
=>
=>
and v_x is the final velocity of the package which is equivalent to the initial velocity ![u_x](https://tex.z-dn.net/?f=u_x)
So
![\beta = tan ^{-1}[-130.05}{57.96 } ]](https://tex.z-dn.net/?f=%5Cbeta%20%20%3D%20%20tan%20%5E%7B-1%7D%5B-130.05%7D%7B57.96%20%7D%20%5D)
![\beta =- 65.55^o](https://tex.z-dn.net/?f=%5Cbeta%20%3D-%2065.55%5Eo)
The negative direction show that it is moving towards the south east direction
Answer:
Lindsey biked 45 miles for 3 hours at 15 mph and walked 8 miles for 2 hours at 4 mph.
Explanation:
Speed = distance/time
Let the distance that Lindsey biked through be x miles and the time it took her to bike through that distance be t hours
Then, the rest of the distance that she walked is (53 - x) miles
And the time she spent walking that distance = (5 - t) hours
Her biking speed = 15 mph = 15 miles/hour
Speed = distance/time
15 = x/t
x = 15 t (eqn 1)
Her walking speed = 4 mph = 4 miles/hour
4 = (53 - x)/(5 - t)
53 - x = 4 (5 - t)
53 - x = 20 - 4t (eqn 2)
Substitute for X in (eqn 2)
53 - 15t = 20 - 4t
15t - 4t = 53 - 20
11t = 33
t = 3 hours
x = 15t = 15 × 3 = 45 miles.
(53 - x) = 53 - 45 = 8 miles
(5 - t) = 5 - 3 = 2 hours
So, it becomes evident that Lindsey biked 45 miles for 3 hours at 15 mph and walked 8 miles for 2 hours at 4 mph.
He goes 24 miles. hope this helps!