Answer:
Explanation:
1. Fall time
i) Find the vertical speed of the plane when the tanks were released
[tex]V_{y,0}=84m/s\times sin(30\º)=42m/s[/tex]
That is the same initial vertical speed of the tanks.
ii) Find the fall time
[tex]y-y_0=V_{y,0}\cdot t+g\cdot t^2/2[/tex]
[tex]120=42t+4.9t^2[/tex]
[tex]4.9t^2+42t-120=0[/tex]
[tex]t=\dfrac{-42\pm\sqrt{(42)^2-4(4.9)(-120)}}{2\times 4.9}[/tex]
Only the positive value has aphysical meaning: t = 2.26 seconds.
2. Speed when they hit the ground
i) The horizontal speed is constant:
[tex]V_x=84m/s\times cos(30\º)\approx72.5m/s[/tex]
ii) The vertical speed is:
[tex]V_y=V_{y,0}+g\cdot t[/tex]
[tex]V_y=42m/s+9.8m/s^2\cdot (2.26s)\approx64.1m/s[/tex]
iii) Total speed
[tex]V=\sqrt{V_x^2+V_y^2}[/tex]
[tex]V=\sqrt{(72.5m/s)^2+(64.1m/s)^2}\approx97m/s[/tex]
