Question

14. 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 0.3 m, find the angle that the stream makes with the vertical as it strikes the ground. (The horizontal distance frorm the bottom of the cylindrical stand to the splash point is unknown.)

Answer 1

By Torricelli's equation

$v = \sqrt{2g\Delta h}$

V2 * 9.81 * 0.3

2.426m/s u

Now this jet which is launched horizontally will start moving under the influence of gravitational acceleration, thereby moving in a parabolic path,
The horizontal component of velocity will remain fixed, as there is no acceleration along the horizontal axis

Consider vertical component of motion

u-02 + 2 * 9.81 * 1 2

し,2 = 4.429m /s

Angle,

an 0

tana 2.426 4.429

with the vertical

θ 28.710