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?

[tex]\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[/tex]

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

Lanuel Lanuel · Answer 2 · 2021-10-24T01:56:28+02:00

If a person standing at the edge of a seaside cliff kicks a stone over the edge, the speed with which the stone strike the water is 36.65 m/s.

Given the following data:

Initial horizontal velocity = 18 m/s (since the stone was kicked over the edge).

Vertical displacement = 52 meters

Initial vertical velocity = 0 m/s

We know that acceleration due to gravity (a) is equal to 9.8 meter per seconds square.

First of all, we would determine the vertical component of the velocity by using the third equation of motion;

[tex]V^2 = U^2 + 2aS[/tex]

Where:

V is the final speed.

U is the initial speed.

a is the acceleration.

S is the displacement.

Substituting the given parameters into the formula, we have;

[tex]V^2 = 0^2 + 2(9.8)(52)\\\\V^2 = 0 + 1038.2\\\\V = \sqrt{1019.2}[/tex]

Vertical component of velocity = 31.92 m/s

To find the speed with which the stone strike the water, we would calculate the resultant speed by doing a vector addition of the horizontal component of the velocity and the vertical component of the velocity;

[tex]Speed = \sqrt{H^2 + V^2} \\\\Speed = \sqrt{18^2 + 31.92^2}\\\\Speed = \sqrt{324 + 1019.2}\\\\Speed = \sqrt{1343.2}[/tex]

Speed = 36.65 m/s

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