Answer:
[tex]\theta = 58.98[/tex]°
Explanation:
given data:
h = 18 ft = 5.48 m
from figure
[tex]h_{max} = 5.48 - 0.50 = 4.98 m[/tex]
[tex]h_{max} = \frac{ v_y^2}{2g}[/tex]
[tex]v_y =\sqrt{2gh_{max}[/tex]
[tex]v_y = \sqrt{2*9.8* 4.98} m/s[/tex]
[tex]v_y = 9.88 m/s[/tex]
[tex]t = \frac{v_y}{g} =\frac{9.88}{9.8} = 1.01 s[/tex]
[tex]v_x = \frac{d}{t} = \frac{6}{1.01} = 5.49 m/s
[tex]v = \sqrt{v_x^2+v_y^2} = \sqrt{5.95^2+9.88^2}[/tex]
v = 11.53 m/s
[tex]tan\theta = \frac[v_x}{v_y}[/tex]
[tex]\theta = tan^{-1} \frac{9.88}{5.94}[/tex]
[tex]\theta = 58.98[/tex]°
