# Answer to Question #16286 in Mechanics | Relativity for Alexandra

Question #16286

A person standing at the edge of a cliff kicks a stone over the edge with a speed of 18m/s . The cliff is 52m above the water. How long does it take for the stone to fall to the water? With what speed does it strike the water?

Expert's answer

The initial horizontal component of velocity doesn't affect the falling time. We know that

H = gT²/2,

and therefore,

T = √(2H/g) = √(2*52[m]/9.8[m/s²]) ≈ 3.3[s].

The vertical speed component of a stone at the moment of landing will be

V(vertical) = gT = 9.8[m/s²]*3.3[s] = 32.34[m/s]

and the magnitude of velocity

V = √((V(vertical))² + (V(horizontal))²) = √((18[m/s])² + (32.34[m/s])²) = 37.0[m/s].

So, it will take about 3.3 seconds the stone to fall down and it will strike water with the speed of 37.0 m/s.

H = gT²/2,

and therefore,

T = √(2H/g) = √(2*52[m]/9.8[m/s²]) ≈ 3.3[s].

The vertical speed component of a stone at the moment of landing will be

V(vertical) = gT = 9.8[m/s²]*3.3[s] = 32.34[m/s]

and the magnitude of velocity

V = √((V(vertical))² + (V(horizontal))²) = √((18[m/s])² + (32.34[m/s])²) = 37.0[m/s].

So, it will take about 3.3 seconds the stone to fall down and it will strike water with the speed of 37.0 m/s.

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