Question

a plane is flying 120 m above the ground at an angle of 30 degrees to the horizontal, when the pilot released 2 fuel tanks to decrease the planes load. How long did the tanks fall and with what speed did it hit the ground of the plane's speed was 84 m/s?

Answer 1

Answer:

• 2.26 seconds

• 97m/s

Explanation:

1. Fall time

i) Find the vertical speed of the plane when the tanks were released

         $V_{y,0}=84m/s\times sin(30\º)=42m/s$

That is the same initial vertical speed of the tanks.

ii) Find the fall time

         $y-y_0=V_{y,0}\cdot t+g\cdot t^2/2$

         $120=42t+4.9t^2$

         

         $4.9t^2+42t-120=0$

         

        $t=\dfrac{-42\pm\sqrt{(42)^2-4(4.9)(-120)}}{2\times 4.9}$

Only the positive value has aphysical meaning: t = 2.26 seconds.

2. Speed when they hit the ground

i) The horizontal speed is constant:

          $V_x=84m/s\times cos(30\º)\approx72.5m/s$

ii) The vertical speed is:

            $V_y=V_{y,0}+g\cdot t$

            $V_y=42m/s+9.8m/s^2\cdot (2.26s)\approx64.1m/s$

iii) Total speed

          $V=\sqrt{V_x^2+V_y^2}$

          $V=\sqrt{(72.5m/s)^2+(64.1m/s)^2}\approx97m/s$