Answer:
velocity is 4.995 m/s
and height = 2.549 mm
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
[tex]\frac{d}{df} mv[/tex] = 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 = [tex]\sqrt{\frac{Mg}{\rho A} }[/tex] .............................1
so put here value
Vf = [tex]\sqrt{\frac{0.2*9.8}{1000*\pi (5*10^{-3})^2} }[/tex]
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 = [tex]\frac{Vi^2-Vf^2}{2g}[/tex]
h = [tex]\frac{5^2-4.995^2}{2g}[/tex]
h = 2.549 mm
