The water jet’s horizontal distance \( D \) from the hole to the point where it strikes the ground depends on the velocity of the jet at height \( h_1 \) and the gravitational pull. The velocity at the hole is given by:
\[ v = \sqrt{2gh_1} \]
This is the velocity of the jet when it leaves the hole at height \( h_1 \), as the velocity is due to the gravitational potential energy being converted into kinetic energy.
The maximum horizontal distance \( D \) occurs when the jet’s flight time and horizontal speed are maximized.
The maximum horizontal distance \( D \) is greatest when the height of the hole is maximized, i.e., \( h_1 = h \). Therefore, the maximum value of \( D \) is:
\[ D = h \]
Thus, the maximum value of \( D \) is indeed \( h \).
In order to achieve the static equilibrium of the see-saw about the fulcrum \( P \), shown in the figure, the weight of Box B should be _________ kg, if the weight of Box A is 50 kg.
A particle of mass 1kg, initially at rest, starts sliding down from the top of a frictionless inclined plane of angle \(\frac{𝜋}{6}\)\(\frac{\pi}{6}\) (as schematically shown in the figure). The magnitude of the torque on the particle about the point O after a time 2seconds is ______N-m. (Rounded off to nearest integer)
(Take g = 10m/s2)