given that snow is projected at an angle of 40 degree
It range is given as a = 19 ft
[tex]a = 19 * 0.3048 = 5.8 m[/tex]
now we can use the formula of horizontal range
[tex]R = \frac{v_o^2 sin2\theta}{g}[/tex]
[tex]5.8 =\frac{ v_o^2 sin(2*40)}{9.8}[/tex]
[tex]5.8 = \frac{v_o^2 * sin80}{9.8}[/tex]
[tex]v_o^2 = 57.7[/tex]
[tex]v_o = 7.6 m/s[/tex]
so its initial speed must be 7.6 m/s