Respuesta :
Given data
(14)
*The given constant velocity of the plane take off is v = 175 km/ h = 48.61 m/s
*The given angle is
[tex]\theta=35.5^0[/tex]*The given time is t = 48.0 s
The vertical component of the velocity is calculated as
[tex]\begin{gathered} v_y=v_{}\sin \theta-gt \\ =(48.61)\times\sin 35.5-(9.8)(48.0) \\ =-442.17\text{ m/s} \end{gathered}[/tex]The horizontal component of the velocity is calculated as
[tex]\begin{gathered} v_x=v\cos \theta \\ =(48.61)\times\cos 35.5^0 \\ =39.57\text{ m/s} \end{gathered}[/tex]The formula for the velocity of the plane is given as
[tex]v=\sqrt[]{(v_x_{})^2+(v_y)^2_{}}[/tex]Substitute the known values in the above expression as
[tex]\begin{gathered} v=\sqrt[]{(39.57)^2+(-442.17)^2} \\ =443.93\text{ m/s} \end{gathered}[/tex]Hence, the velocity of the plane in meter per second (m/s) is v = 443.93 m/s
