Question

Two water jets are emerging from a vessel at a height of 50 centimeters and 100 centimeters. If their horizontal velocities at the point of ejection are 1 meter/second and 0.5 meters/second respectively, calculate the ratio of their horizontal distances of impact.
A. 1:04 B. 1:234 C. 2:01 D. 1.41:1 E. 3:02

Answer 1

Answer: The correct answer is option (D).

Explanation:

Vertical distance  of first water jet = $y_1=50 cm$

Vertical distance of second water jet = $y_2=100 cm$

Horizontal velocity of first water jet =$u_y_ 1=1 m/s$

Horizontal velocity of second water jet = $u_y_2=0.5 m/s$

Vertical distance of first water jet:

$y_1=(u_y_1)t_1+\frac{1}{2}gt_1^2$

$u_y_1=0,(u_y_1)t_1=0$

$t_1=\sqrt{\frac{2y_1}{g}}$...(1)

Similarly, Vertical distance of second water jet:

$y_2=(u_y_2)t_2+\frac{1}{2}gt_2^2$

$u_y_2=0,(u_y_2)t_2=0$

$t_2=\sqrt{\frac{2y_2}{g}}$...(2)

Horizontal distance of first water jet:

$x_1=(u_x_1)t_1+\frac{1}{2}at_1^2$

There is no acceleration in horizontal direction.

$a=0,at_1^2=0$

$x_1=(u_y_1)t_1$...(3)

Horizontal distance of second water jet:

$x_2=(u_x_2)t_2+\frac{1}{2}at_2^2$

$a=0,at_2^2=0$

$x_2=(u_x_2)t_2$...(4)

On dividing (3) and (4):

$\frac{x_1}{x_2}=\frac{(u_y_1)t_1}{(u_y_2)t_2}$

$\frac{x_1}{x_2}=\frac{u_y_1}{u_y_2}\times\frac{\sqrt{\frac{2y_1}{g}}}{\sqrt{\frac{2y_2}{g}}}}=\frac{u_y_1}{u_y_2}\sqrt{\frac{y_1}{y_2}}$

(from (1) and (2))

$\frac{x_1}{x_2}=\frac{1 m/s}{0.5 m/s}\sqrt{\frac{50 cm}{100 cm}}=1.41:1$

Hence, the correct answer is option (D).