Anthony Pitt
Mar 5
Abc?
The path of a jet of water coming from the opening in a side of a water tank is such that the jet falls through a distance of 120mm in a horizontal distance of 1.2 m. What is the velocity ( assumed horizontal) with which the jet leaves the tank?
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ChatGPT
Mar 5
To find the velocity with which the jet leaves the tank, we can use the equation of motion for projectile motion:
y = (1/2)gt^2
Where:
y = vertical distance fallen (120mm = 0.12m)
g = acceleration due to gravity (9.81 m/s^2)
t = time taken for the jet to fall
We can rearrange the equation to solve for time:
0.12 = (1/2)(9.81)t^2
0.24 = 9.81t^2
t^2 = 0.0245
t = √0.0245
t ≈ 0.1567 seconds
Next, we can use the horizontal distance and time to find the horizontal velocity:
v = d/t
v = 1.2 / 0.1567
v ≈ 7.65 m/s
Therefore, the velocity with which the jet leaves the tank is approximately 7.65 m/s.
y = (1/2)gt^2
Where:
y = vertical distance fallen (120mm = 0.12m)
g = acceleration due to gravity (9.81 m/s^2)
t = time taken for the jet to fall
We can rearrange the equation to solve for time:
0.12 = (1/2)(9.81)t^2
0.24 = 9.81t^2
t^2 = 0.0245
t = √0.0245
t ≈ 0.1567 seconds
Next, we can use the horizontal distance and time to find the horizontal velocity:
v = d/t
v = 1.2 / 0.1567
v ≈ 7.65 m/s
Therefore, the velocity with which the jet leaves the tank is approximately 7.65 m/s.
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