Question

A jet of water squirts out horizontally from a hole
near the bottom of the very large tank in the figure.
If the height, h, of the water level in the tank is 0.5 m,
find the angle that the stream makes with the vertical
as it strikes the ground. (The horizontal distance from
the bottom of the cylindrical stand to the splash point
is unknown.)

Answer 1

Initial height of the jet of water = h = 0.5 m.

Hence, from Bernoulli's principle, horizontal velocity = v_x = 2gh,

where, g = 9.8 m / s² is gravitational acceleration.

Hence, v_x = ( 2 x 9.8 x 0.5 ) m / s ~ 3.13 m / s.

Initial vertical velocity = 0 m / s.

Hence, final vertical velocity v_y is given by : v_y² = 0² + 2gh,

or, v_y = ( 2 x 9.8 x 0.5 ) m / s ~ 3.13 m / s.

Hence, the angle that the stream makes with the vertical as it strikes the ground :

tan^-1 ( v_y / v_x ) = tan^-1 ( 3.13 / 3.13 ) = 45^o.