Respuesta :
Answer: The correct answer is option (D).
Explanation:
Vertical distance of first water jet = [tex]y_1=50 cm[/tex]
Vertical distance of second water jet = [tex]y_2=100 cm[/tex]
Horizontal velocity of first water jet =[tex]u_y_ 1=1 m/s[/tex]
Horizontal velocity of second water jet = [tex]u_y_2=0.5 m/s[/tex]
Vertical distance of first water jet:
[tex]y_1=(u_y_1)t_1+\frac{1}{2}gt_1^2[/tex]
[tex]u_y_1=0,(u_y_1)t_1=0[/tex]
[tex]t_1=\sqrt{\frac{2y_1}{g}}[/tex]...(1)
Similarly, Vertical distance of second water jet:
[tex]y_2=(u_y_2)t_2+\frac{1}{2}gt_2^2[/tex]
[tex]u_y_2=0,(u_y_2)t_2=0[/tex]
[tex]t_2=\sqrt{\frac{2y_2}{g}}[/tex]...(2)
Horizontal distance of first water jet:
[tex]x_1=(u_x_1)t_1+\frac{1}{2}at_1^2[/tex]
There is no acceleration in horizontal direction.
[tex]a=0,at_1^2=0[/tex]
[tex]x_1=(u_y_1)t_1[/tex]...(3)
Horizontal distance of second water jet:
[tex]x_2=(u_x_2)t_2+\frac{1}{2}at_2^2[/tex]
[tex]a=0,at_2^2=0[/tex]
[tex]x_2=(u_x_2)t_2[/tex]...(4)
On dividing (3) and (4):
[tex]\frac{x_1}{x_2}=\frac{(u_y_1)t_1}{(u_y_2)t_2}[/tex]
[tex]\frac{x_1}{x_2}=\frac{u_y_1}{u_y_2}\times\frac{\sqrt{\frac{2y_1}{g}}}{\sqrt{\frac{2y_2}{g}}}}=\frac{u_y_1}{u_y_2}\sqrt{\frac{y_1}{y_2}}[/tex]
(from (1) and (2))
[tex]\frac{x_1}{x_2}=\frac{1 m/s}{0.5 m/s}\sqrt{\frac{50 cm}{100 cm}}=1.41:1[/tex]
Hence, the correct answer is option (D).
