projectile motion

Using conservation of energy, consider a mass m of water,
kinetic energy of water before the fall = (1/2)m.(4^2) J = 8m J
potential energy of water = mg.(12) = 120m J
[where g is the acceleration due to gravity, taken to be 10 m/s2]
hence, total energy = (8m + 120m) J = 128m J

Kinetic energy after the fall = (1/2)mv^2
where v is the speed after the fall
hence, (1/2)mv^2 = 128m
solve for v gives v = 16 m/s

Althernatively, you may use the equation of motion: v^2 = u^2 + 2a.s and apply it to the vertical fall of water,
v^2 = 0^2 + 2.(-10).(-12)
i.e. v = 15.492 m/s

Hence, the speed after the fall = square-root[15.492^2 + 4^2] m/s = 16 m/s

A river flows over a waterfall which is 12 m high. If the water flows horizontally at 4 ms^-1 just before the fall. calculate its speed when it hits the bottom.

Suppose air resistance is neglected and acceleration of gravity 10ms^-2.

Consider a drop of water with mass m,

its vertical speed when it hits the bottom can be found by using P.E. = K.E. equation:

mg x 12 = 1/2 m v^2

v = SQRT(240)

At the same time,
its horizontal speed keeps 4ms^-1, denoted as u.

Therefore,

combining horizontal and vertical speed vectors by using pyth thm, we get

the speed of the water = SQRT(u^2+v^2)
= SQRT(16+240)
= 16 (ms^-1)