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

• 2.26 seconds

• 97m/s

Step-by-step explanation:

1. Fall time

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

`V_(y,0)=84m/s* 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=(-42\pm√((42)^2-4(4.9)(-120)))/(2* 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* 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=√(V_x^2+V_y^2)`

`V=√((72.5m/s)^2+(64.1m/s)^2)\approx97m/s`