Answer:
63.6 m/s
Explanation:
The motion of the water inside the waterfall is the motion of the projectile, so it has:
- a horizontal velocity which is constant: [tex]v_x = 2.90 m/s[/tex], since there are no forces acting along the horizontal direction
- a vertical motion which is an accelerated motion with constant acceleration, g=9.8 m/s^2, due to the effect of the force of gravity
The magnitude of the vertical velocity of the water when it hits the ground can be found by using
[tex]v_y^2 - u_y^2 = 2gh[/tex]
where
vy is the final velocity
uy=0 is the initial velocity
h = 206 m is the height of the waterfall
Solving for vy,
[tex]v_y = \sqrt{2gh}=\sqrt{2(9.8 m/s^2)(206 m)}=63.5 m/s[/tex]
so this is the vertical velocity of the water as it hits the ground.
The magnitude of the overall velocity will be the resultant of the horizontal and the vertical components:
[tex]v=\sqrt{v_x^2+v_y^2}=\sqrt{(2.90 m/s)^2+(63.5 m/s)^2}=63.6 m/s[/tex]
