Answer to Question #16286 in Mechanics | Relativity for Alexandra

Question

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?

Answer 1

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.