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. 1:04 1:234 2:01 1.41:1 3:02

For t1:

t1 = square root of 2h1 / g = square root of 2 * 0.5 / 9.8 = 0.319 sec

For t2:

t2 = sqaure root of 2h2 / g = square root of 2 * 1.0 / 9.8 = 0.451 sec

Wherein:
t = time(s) for the vertical movement
h= height
g = gravity (using the standard 9.8 m/sec measurement)

d1 = 1*0.319 = 0.319 m
d2 = 0.5 * 0.451 = 0.225 m

Where:

d = hor. distance

ratio = d1:d2
= 0.319 : 0.225
=3.19 : 2.25

The answer is 3.19 : 2.25

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aristocles aristocles · Answer 2 · 2019-07-01T03:33:03+02:00

Answer:

1.41 : 1

Explanation:

time required to reach the ground is given as

[tex]t = \sqrt{\frac{2h}{g}}[/tex]

now we know that

h = 50 cm and h = 100 cm

so the ratio of time of two is given as

[tex]\frac{t_1}{t_2} = \sqrt{\frac{50}{100}}[/tex]

[tex]\frac{t_1}{t_2} = \sqrt{\frac{1}{2}}[/tex]

now the distance traveled by the object is given as

[tex]d = vt[/tex]

now the ratio of the distance moved is given as

[tex]\frac{d_1}{d_2} = \frac{v_1t_1}{v_2t_2}[/tex]

[tex]\frac{d_1}{d_2} = \frac{1}{0.5}\times \sqrt{\frac{1}{2}}[/tex]

[tex]\frac{d_1}{d_2} = \sqrt 2 = 1.41 : 1[/tex]