Question

An airplane with a speed of 92.3 m/s is climbing upward at an angle of 51.1 ° with respect to the horizontal. When the plane's altitude is 532 m, the pilot releases a package.
(a) Calculate the distance along the ground, measured from a point directly beneath the point of release, to where the package hits the earth.
(b) Relative to the ground, determine the angle of the velocity vector of the package just before impact. (a) Number Units (b) Number Units

Answer 1

Answer:

a

$D = 1162.7 \ m$

b

$\beta =- 65.55^o$

Explanation:

From the question we are told that

  The speed of the airplane is  $u = 92.3 \ m/s$

   The  angle is  $\theta = 51.1^o$

    The altitude of the plane is  $d = 532 \ m$

Generally the y-component of the airplanes velocity is  

       $u_y = v * sin (\theta )$

=>     $u_y = 92.3 * sin ( 51.1 )$

=>     $u_y = 71.83 \ m/s$

Generally the displacement  traveled by the package in the vertical direction is

       $d = (u_y)t + \frac{1}{2}(-g)t^2$

=>       $-532 = 71.83 t + \frac{1}{2}(-9.8)t^2$

Here the negative sign for the distance show that the direction is along the negative y-axis

 =>   $4.9t^2 - 71.83t - 532 = 0$

Solving this using quadratic formula we obtain that

    $t = 20.06 \ s$

Generally the x-component of the velocity is  

     $u_x = u * cos (\theta)$

=>    $u_x = 92.3 * cos (51.1)$

=>   $u_x = 57.96 \ m/s$

Generally the distance travel in the horizontal  direction is    

     $D = u_x * t$

=>   $D = 57.96 * 20.06$

=>    $D = 1162.7 \ m$

Generally the angle of the velocity vector relative to the ground is mathematically represented as

       $\beta = tan ^{-1}[\frac{v_y}{v_x } ]$

Here $v_y$ is the final  velocity of the package along the vertical  axis and this is mathematically represented as  

     $v_y = u_y - gt$

=>  $v_y = 71.83 - 9.8 * 20.06$

=>  $v_y = -130.05 \ m/s$  

and  v_x is the final  velocity of the package which is equivalent to the initial velocity $u_x$

So

       $\beta = tan ^{-1}[-130.05}{57.96 } ]$

       $\beta =- 65.55^o$

The negative direction show that it is moving towards the south east direction