Answer :
a) 36.5 m/s
The motion of the package along the vertical direction is a free fall motion, so we can use the following suvat equation:
[tex]s=u_y t + \frac{1}{2}gt^2[/tex]
where, chosing downward as positive direction:
s = 360 m is the vertical displacement of the package
[tex]u_y[/tex] is the initial vertical velocity of the package, which is equal to that of the plane
t = 6.4 s is the time of flight
[tex]g=9.8 m/s^2[/tex] is the acceleration of gravity
Solving for [tex]u_y[/tex], we find:
[tex]u_y = \frac{s}{t}-\frac{1}{2}gt=\frac{360}{6.4}-\frac{1}{2}(9.8)(6.4)=24.9 m/s[/tex]
The initial vertical velocity is related to the initial speed of the plane by
[tex]u_y = u sin \theta[/tex]
where
u is the initial speed of the plane
[tex]\theta=43^{\circ}[/tex]
Solving for u,
[tex]u=\frac{u_y}{sin \theta}=\frac{24.9}{sin 43}=36.5 m/s[/tex]
b) 170.9 m
The horizontal velocity of the package is given by
[tex]v_x = u cos \theta = (36.5)(cos 43)=26.7 m/s[/tex]
where
u = 36.5 m/s is the initial speed
[tex]\theta=43^{\circ}[/tex]
The horizontal velocity of the package is constant since there are no forces along this direction: therefore, the horizontal distance travelled by the package is given by
[tex]d=v_x t[/tex]
And substituting t = 6.4 s, we find
[tex]d=(26.7)(6.4)=170.9 m[/tex]