Answer:
58.6 degrees below the horizontal
Explanation:
The horizontal velocity of the object remains constant:
[tex]v_x = 2.2 m/s[/tex]
The vertical motion instead is an accelerated motion, so the vertical velocity changes according to
[tex]v_y ^2 - u_y^2 = 2gd[/tex]
where
[tex]v_y[/tex] is the final vertical velocity
[tex]u_y = 0[/tex] is the initial vertical velocity
g = -9.8 m/s^2 is the acceleration of gravity (downward, so with a negative sign)
d = -0.65 m is the vertical displacement of the object
Solving the equation for [tex]v_y[/tex],
[tex]v_y = \sqrt{u_y^2 +2gd}=\sqrt{0+2(-9.8 m/s^2)(-0.65 m)}=3.6 m/s[/tex]
The vertical velocity just before the impact is therefore -3.6 m/s (negative since it points downward).
The angle of impact below the horizontal therefore will be
[tex]\theta=tan^{-1} (\frac{v_y}{v_x})=tan^{-1} (\frac{3.6 m/s}{2.2 m/s})=58.6^{\circ}[/tex]
below the horizontal.
