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.

Answer 1

The ratio of the horizontal distances of impact for the two water jets is 2.

Step-by-step explanation:

To calculate the ratio of the horizontal distances of impact, we need to first find the time of flight for each water jet. The time of flight can be determined using the formula:

time = distance / velocity

For the first water jet with a height of 50 centimeters and a velocity of 1 meter/second, the time of flight is:

time 1 = 0.5 meters / 1 meter/second

= 0.5 seconds

For the second water jet with a height of 100 centimeters and a velocity of 0.5 meter/second, the time of flight is:

time 2 = 1 meter / 0.5 meter/second

= 2 seconds

Now, we can calculate the horizontal distances of impact using the formula:

distance = velocity * time

For the first water jet, the horizontal distance of impact is:

distance 1 = 1 meter/second * 0.5 seconds

= 0.5 meters

For the second water jet, the horizontal distance of impact is:

distance2 = 0.5 meter/second * 2 seconds = 1 meter

Therefore, the ratio of their horizontal distances of impact is:

ratio = distance 2 / distance 1

= 1 meter / 0.5 meters

= 2

Answer 2

t1 = √2h1/g = √2*0.5/9,8 = 0.319 sec
t2 = √2h2/g = √2*1.0/9,8 = 0.451 sec

In which t = times for the vertical movement
h = height
g= gravity (we use standardized measurement of 9.8)

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

in which d = Horizontal distance

ratio
= di : d2
= 0.319 : 0.225

= 3.19 : 2.25