Question

A 0.20kg ball is held stationary by a jet of water from falling. At the nozzle the speed of the jet is 5m/s, and its diameter is 10mm. find the velocity at which the water jet hits the ball and the height of the ball. Density of water is 1000kg/m^3. Assume theres no change in diameter of the water jet. (a)Find the velocity of the water jet as it hits the ball.(hint: the weight of the ball is held up by the force of water jet) (b)-find the height at which the ball is held

Answer 1

velocity is 4.995 m/s

and height = 2.549 mm

Step-by-step explanation:

given data

mass = 0.20 kg

speed = 5 m/s

diameter = 10 mm

density = 1000 kg/m³

to find out

find the velocity at which the water jet hits the ball and the height of the ball

solution

we know here that force applied by jet is

`(d)/(df) mv` = M×Vf

and Force = ρ ×A×Vf²

here ρ is density A is area and Vf is velocity and M is mass

so as per free body diagram

force F = Mg

so

velocity will be

Vf =
`\sqrt{(Mg)/(\rho A) }` .............................1

so put here value

Vf =
`\sqrt{(0.2*9.8)/(1000*\pi (5*10^(-3))^2) }`

Vf = 4.995

so velocity is 4.995 m/s

and

as per diagram we apply bernoulli equation between x and y

P(x) + 0.5 ρ Vi² + ρgh(x) = P(y) + 0.5 ρ Vf² + ρgh(y)

ρgh = 0.5 ρ ( Vi² - Vf²)

h =
`(Vi^2-Vf^2)/(2g)`

h =
`(5^2-4.995^2)/(2g)`

h = 2.549 mm