Answer
given,
eagle horizontal velocity = 5.5 m/s
resultant speed is given by
[tex]v = \sqrt{v_x^2+v_y^2}[/tex]
[tex]v_y = \sqrt{v^2-v_x^2}[/tex]
[tex]v_y = \sqrt{(2\times 5.5)^2-5.5^2}[/tex]
[tex]v_y = 9.53\ m/s[/tex]
[tex]t = \dfrac{v_y}{g} [/tex]
[tex]t = \dfrac{9.53}{9.81}[/tex]
t = 0.971 s
b)
resultant speed is given by
[tex]v = \sqrt{v_x^2+v_y^2}[/tex]
[tex]v_y = \sqrt{v^2-v_x^2}[/tex]
[tex]v_y = \sqrt{(2\times 9.53)^2-5.5^2}[/tex]
[tex]v_y = 18.25\ m/s[/tex]
[tex]t' = \dfrac{v_y}{g} [/tex]
[tex]t' = \dfrac{18.25}{9.81}[/tex]
t' = 1.86 s
therefore additional time is equal to = t' - t
= 1.86 - 0.971
= 0.889 s
