Question

a person standing at the edge of a seaside cliff kicks a stone over the edge with a speed of 18 m/s. The cliff is 52 m above the water's surface. With what speed did it strike the water?

Answer 1

Answer:

Speed with which it strike the water = 36.66 m/s

Explanation:

Horizontal component of velocity = 18 m/s

We need to find vertical component of velocity.

Considering vertical motion of stone

Initial velocity, u =  0 m/s

Acceleration , a = 9.81 m/s²

Displacement, s = 52 m

We have equation of motion v² = u² + 2as

Substituting

   v² = u² + 2as

    v² = 0² + 2 x 9.81 x 52

    v = 31.94 m/s

Vertical component of velocity = 31.94 m/s

$\texttt{Total velocity = }\\\\\sqrt{\texttt{Horizontal component of velocity}^2+\texttt{Vertical component of velocity}^2}\\\\\texttt{Total velocity = }\sqrt{18^2+31.94^2}=36.66m/s$

Speed with which it strike the water = 36.66 m/s

Answer 2

You already have the speed, now you need the time.
I will use the formula for speed which is S=D/T.
S=Speed    D=Distance   T=Time.
So here we have, 18m/s = 52m/T
we do 18 divided by 52 which would be .3461.
.3461 seconds is how long it took the stone to reach the water.